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NOTRE DAME MATHEMATICAL LECTURES
Number 2
GALOIS THEORY
Lectures delivered at the University of Notre Dame
by
DR. EMIL ARTIN
Professor of Mathematics, Princeton University
Edited and supplemented with a Section on Applications
by
DR. ARTHUR N. MILGRAM
Associate Professor of Mathematics, University of Minnesota
Second Edition
With Additions and Revisions
UNIVERSITY OF
NOTRE DAME
NOTRE
DAME PRESS
LONDON
Copyright 1942, 1944
UNNERSITY OF NOTRE DAME
Second Printing, February 1964
Third Printing, July 1965
Fourth Printing, August 1966
New composition with corrections
Fifth Printing, March 1970
Sixth Printing, January 197 1
Printed in the United States of America by
NAPCO Graphic Arts, Inc., Milwaukee, Wisconsin
TABLE OF CONTENTS
(The sections marked with an asterisk
have been herein added to the content
of the first edition)
Page
LINEAR
A
ALGEBRA ................................... .
Fields ......................................... .
Vector Spaces ................................... .
C. Homogeneous Linear Equations . . . . . . . . . . . . . . . . . . . . .
D.
Dependence and Independence of Vectors .. , . . . . . . . . .
E. Non-homogeneous Linear Equations . . . . . . . . . . . . . . . . .
F.* Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
B.
II
FIELD
THEORY ............................ .
A.
B.
C.
D.
E.
Extension Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Algebraic Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Unique Decomposition of Polynomials
into hreducible Factors . . . . . . . . . . . . . . . . . . . . .
F. Group Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
G.* Applications and Examples to Theorem 13 . . . . . . . . . . . .
H, Nmmal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
I.
Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
].
Roots of Unity.............................
.., ..
K. Noether Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
L
Kummer's Fields . . . . . . . . . . . . . . . . . . . . . . .
..........
M. Simple Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
N. Existence of a Normal Basis . . . . . . . . . . . , . . . . . . . . . . .
Q, Theorem on Natural hrationalities . . . . . . . . . . . . . . . . . . .
Ill
APPLICATIONS
By A. N. Milgram •......................
A. Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
B. Permutation Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
C. Solution of Equations by Radicals . . . . . . . . . . . . . . . . . . .
D. The General Equation of Degree n. . . . . . . . . . . . . . . . . . .
E. Solvable Equations of Prime Degree . . . . . . . . . . . . . . . . .
F. Ruler and Compass Construction . . . . . . . . . . . . . . . . . . . .
2
4
9
11
21
21
22
25
30
33
34
38
41
49
56
57
59
64
66
67
69
69
70
72
74
76
80
1 LINEAR ALGEBRA
A. FVildB.
(ft·;
c.O\·
f
S)
A field is a set of elements in which a pair of operations called
multiplication and addition is defined analogous to the operations of
multiplication and addition in the real number system (which is itself
an example of a field). In each field F there exist unique elements
called o and 1 which, under the operations of addition and multiplica-
tion, behave with respect to all the other elements ofF exactly as
their correspondents in the real number system. In two respects, the
analogy is not complete:
1) multiplication is not assumed to be commu-
tative in every field, and 2) a field may have only a finite number
of elements.
More exactly, a field is a set of elements which, under the above
mentioned operation of addition, forms an additive abelian group and
for which the elements, exclusive of zero, form a multiplicative group
and, finally, in which the two group operations are connected by the
distributive law. Furthermore, the product of o and any element is defined to be o.
If multiplication in the field is commutative, then the field is
called a commutative field.
B. Vector Spaces.
If Vis an additive abelian group with elements A, B, ... ,
F a field with elements a, b, ... , and if for each a
f
F and A f V
2
the product aA denotes an element of V, then V is called a (left)
vector space over F if the following assumptions hold:
l)a(A+B)=aA+aB
2)
(a + b)A = aA + bA
3) a(bA) = ( ab )A
4) lA = A
The reader may readily verify that if V is a vector space over F, then
oA = 0 and aO = 0 where o is the zero element of F and 0 that of V.
For example, the first relation follows from the equations:
aA
= (a+
o)A
= aA + oA
Sometimes products between elements ofF and V are written in
the form A a in which case V is called a right vector space over F to
distinguish it from the previous case where multiplication by field ele~
ments is from the left. If, in the discussion, left and right vector
spaces do not occur simultaneously, we shall simply use the term
"vector
space."
C. Homogeneous Linear Equations.
If in afield F, aij, i = 1,2, ... , m,j = 1,2, ... , n arem. n ele-
ments, it is frequently necessary to know conditions guaranteeing the
existence of elements in F such that the following equations are satisfied:
(1)
am 1x 1 + am2x 2 + . . . + amnxn = 0.
The reader
will recall that such equations are called linear
homogeneous equations, and a set of elements,
X ,
1
x2 ,
, .. , X
of F, for which all the above equations are true, is called
0
a solution of the system. If not all of the elements x 1 , x , .
xn
2
are o the solution is called non-trivial; otherwise, it is called trivial.
THEOREM 1. A system of linear homogeneous equations always
has a non-trivial solution if the number of unknowns exceeds the number of equations.
The
proof
of this follows the method familiar to most high school
students, namely, successive elimination of unknowns. If no equations
in n > 0 variables are prescribed, then our unknowns are unrestricted
and we may set them all = 1.
We shall proceed by complete induction. Let us suppose that
each
system of k equations in more than k unknowns has a non-trivial
solution when k < m. In the system of equations (1) we assume that
n > m, and
denote
the expression ailx 1 +
We seek elements x , .
1
=
, xn not all o such that L = L =
1
2
If a 1J ~ o for each i and j, then any choice of x ,
1
L
m
. , xn will serve as
a solution. If not all aij are o, then we may assume that a
11
f
o, for
the order in which the equations are written or in which the unknowns
are numbered has no influence on the existence or non-existence of a
simultaneous solution. We can find a non-trivial solution to our given
system of equations, if and only if we can find a non-trivial solution
to the following system:
L1 =
o
L2- a21ai1-1Lt =
0
0.
For, if x 1 ,
L1
, .
,
xn is a solution of these latter equations then, since
= o, the second term in each of the remaining equations is o and,
Lm = o. Conversely, if (1) is satisfied, then
the new system is clearly satisfied. The reader will notice that the
new system was set up in such a way as to "eliminate" x from the
1
last m-1 equations. Furthermore, if a non-trivial solution of the last
m-1 equations, when viewed as equations in x ,.
2
, xn, exists then
+ ~n Xn) would give us a
solution to the whole system. However, the last m-1 equations have
a solution by our inductive assumption, from which the theorem follows.
Remark: If the linear homogeneous equations had been written
in the form ~xja ij:::: o, j :::: 1, 2,
, n, the above theorem would still
hold and with the same proof although with the order in which terms
are written changed in a few instances.
D. Dependence and Independence of Vectors.
In a vector space V over a field F, the vectors A,
called dependent if there exist elements x , .
1
, xn, not
, A n are
all o, ofF such
+ xnAn = 0. If the vectors A,
not dependent, they are called independent.
The dimension of a vector space V over a field F is the maximum
number of independent elements in V. Thus, the dimension of V is n if
there are n independent elements in V, but no set of more than n
independent elements.
A system A,
, A, of elements in V is called a
generating system of V if each element A of V can be expressed
5
m
Am, i.e., A = :£a1.A, for a suitable choice
linearly in terms of A,
i ~ 1
m,inF.
of ai, i = 1,.
THEOREM 2. In any generating system the maximum number of
independent vectors is equal to the dimension of the vector space .
Let A,
. , A,, be a generating system of a vector space V of
dimension n. Let r be the maximum number of independent elements in
the generating system. By a suitable reordering of the generators we may as-
sume A , . . . , Ar independent. By the definition of dimension it follows that
1
r
< n.
For each j, A,
expressing this, a
of A,
r+i
A,.
f
A . are dependent, and in the relation
"'
o, for the contrary would assert the dependence
, Ar. Thus,
It follows that A,
, Ar is also a generating system since in the
linear relation for any element of V the terms involving Ar+i, j I: o, can
all be replaced by linear expressions in A,
Now, let B 1 ,
... ,
Bt be any system of vectors in V where t > r,
then there exist a .. such that B."""" L a .. A., j = 1, 2,
lJ
J
i"""" l
lj
, t, since the
1
A/ s form a generating system. If we can show that B,
dependent, this will give us r _2 n, and the theorem will follow from
this together with the previous inequality r
< n~
Thus, we must ex-
hibit the existence of a non-trivial solution out of F of
the equation
0.
6
'lb this end, it will be sufficient to choose the x.'s so as to satisfy
t
'
the linear equations I
x. a .. = o, i = 1, 2, ... , r, since these ex-
i=l
J
t
lJ
pressions will be the coefficients of A 1 when in I x. B. the BJ 's are
r
j==tJJ
replac;d by i~I a ij A 1 and terms are collected. A solution to the equations
I x.a
..
J
. i=l
= 0, i =
1,2, . . . , r, always exists by Theorem 1.
lj
Remark: Any n independent vectors A, ... , A, in an n dimensional vector space form a generating system. For any vector A, the
vectors A, A, ... , A, are dependent and the coefficient of A, in the
dependence relation, cannot be zero. Solving for A in terms of
A 1'
... ,
A, exhibits A, ... ,An as a generating system.
A subset of a vector space is called a subspace if it is a sub-
group of the vector space and if, in addition, the multiplication of any
element in the subset by any element of the field is also in the subset.
If Ai, ... , As are elements of a vector space V, then the set of allele-
ments of the form a, A, + ... + as As clearly forms a subspace of V.
It is also evident, from the definition of dimension, that the dimension
of any subspace never exceeds the dimension of the whole
vector space.
An s-tuple of elements ( a,, ... , a.) in a field F will be called
a _row vecto_r. The totality of such s-tuples form a vector space if
we define
a) (
a 1 , a2
, ••• ,
a.)= ( b 1 , h 2
, ••• ,
b,) if and only if
a, = b 1' i = 1,... ' s,
{3) (a 1 ,a 2 ,
...
,a,)
1
(b 1 ,h 2 ,
...
,h,)
y) b(a 1 ,a 2
, •••
,a,)= (ba 1 ,ba 2
, ••• ,
ba,), for ban
element of F.
w"~ '"" ''"'~' "" ""'~ -"""''()
they will be called column vectors.
THEOREM 3. The row (column) vector space Fn of all n-tuples
from a field F is a vector space of dimension n over F.
The n elements
(J
,, = ( o, 1,
(n
1
Q)
0 ' .•• '
0)
. . . , 0,
1)
= ( 11 0, 0
= ( O, 0,
1 • ,
are independent and generate Fn. Both remarks follow from the relation
(al,a2, ... ,an)=
2ai(i,
We call a rectangular anay
amt 8 m2'' ' 8 mn
of elements of a field F a matrix. By the right row rank of a matdx, we
mean the maximum number of independent row vectors among the rows
(ail',,., a,) of the matrix when multiplication by field elements is
from the right. Similarly, we define left row rank, light column rank and
left column rank.
THEOREM 4. In any matrix the right column rank equals the left
row rank and the left column rank equals the right row rank. If the field
8
is commutative, these four numbers are equal to each other and are
called the rank of the matrix.
Call the column vectors of the matrix Cl'.
vectors R 1 , .
. , Rm. The column vector 0 is
dependence C 1 x 1 + C 2 x 2 +
. , Cn and the row
and any
0 is equivalent to a
solution of the equations
0
(1)
Any change in the order in which the rows of the matrix are written
gives
rise
to the same system of equations and,
hence, does
not change
the column rank of the matrix, but also does not change the row rank
since the changed matrix would have the same set of row vectors. Call
c the right column rank and r the left row rank of the matrix. By the
above remarks we may assume that the first r rows are independent row
vectors. The row vector space generated by
all
the rows of the matrix
has, by Theorem 1, the dimension r and is even generated by the first
r rows. Thus, each row after the
the first r rows. Consequently,
rth
IS linearly expressible in terms of
any solution of the first r equations in
(1) will be a solution of the entire system since any of the last n-r
equations is obtainable as a linear combination of the first r.
Con-
versely, any solution of (1) will also be a solution of the first r
equations. This means that the matrix
9
allal2. ·a,n)
(
8 rt 8 r2 ·
arn
consisting of the first r rows of the original matrix has the same right
column rank as the original. It has also the same left row rank since
the r rows were chosen independent. But the column rank of the amputated matrix car-mot exceed r by Theorem 3. Hence, c
calling
c'
~ r.
Similarly,
the left column rank and r' the right row rank, c'
<... r'.
If we form the transpose of the original matrix, that is, replace rows by
columns and columns by rows, then the left row rank of the transposed
matrix equals the left column rank of the original. If then to the
transposed matrix we apply the above
considerations
we arrive at
r < c and r' < c'.
E. Non-homogeneous Linear Equations.
The system of non-homogeneous linear equations
3 11 xl + 3 12x2 +
a21 xl +
+
8 1nxn =
bl
+ a2nxn =
b2
(2)
has a solution if and only if the column vector
in the space generated by the vectors
cr
Cf(U
10
This means that there is a solution if and only if the right column rank of
the matrix ( a 11 ..• a 1" ) is the same as the
amt· .. amw
right column rank of the augmented matrix ( : ' ·
~n b1
)
ami* '' 8 mnbm
since the vector space generated by the original must be the same as
the vector space generated by the augmented matrix and in either case
the dimension is the same as the rank of the matrix by Theorem 2.
By Theorem 4, this means that the row tanks are equal. Conversely, if the row rank of the augmented matrix is the same as the row
rank of the original matdx, the column ranks will be the same and the
equations will have a solution.
If the equations (2) have a solution, then any relation among the
rows of the original matrix subsists among the rows of the augmented
matrix. For equations (2) this merely means that like combinations
of equals are equal. Conversely, if
each relation which subsists be-
tween the rows of the original matrix also subsists between the rows
of the augmented matrix, then the row rank of the augmented matrix
is the same as the row rank of the original matrix. In terms of the
equations this means that there will exist a solution if and only if
the equations are consistent, i.e., if and only if any dependence
between the left hand sides of the equations also holds between the
light sides.
11
THEOREM 5. If in equations (2) m = n, there exists a unique
solution if and only if the conesponding homogeneous equations
allxt + ~2x2 + . . . + alnxn
8 n1 xl
+ an2x2 + . . . + 3 nnxn
= 0
°
have only the trivial solution.
If they have only the trivial solution, then the column vectors
are independent. It follows that the original n equations in n unknowns
will have a unique solution if they have any solution, since the difference, term by term, of two distinct solutions would be a non-trivial
solution of the homogeneous equations. A solution would exist since
the n independent column vectors form a generating system for the
n-dimensional space of column vectors.
Conversely, let us suppose our equations have one and only one
solution. In this case, the homogeneous equations added term by
te1m
to a solution of the original equations would yield a new solu-
tion to the original equations. Hence, the homogeneous equations have
only the trivial solution.
F. Detenninants. n
The theory of determinants that we shall develop in this chapter
is not needed in Galois theory. The reader may 1 therefore, omit this
section if he so desires.
We assume our field to be c o m m ut a t i v e and consider the
square matrix
------~~--~~----~~~~~-----
1) Of the preceding theory only Theorem 1, for
homogeneous equations and the notion of
linear dependence are assumed known.
12
(1)
ani an2 . . . . 3 nn
of n rows and n columns. We shall define a certain function of this
matrix whose value is an element of our field. The function
will
be
called the determinant and will be denoted by
8..2ta22···· 3 2n
(2)
'"'''''""
or by D ( A , A ,. , , An) if we wish to consider it as a function of the
1
2
column vectors A, A,
. A, of (1). If we keep all the columns but A,
constant and consider the determinant as a function of A, then we
write Dk( Ak) and sometimes even only D.
Definition. A function of the column vectors is a determinant if
it satisfies the following three axioms:
1. Viewed as a function of any column A, it is linear and homogeneous, i.e . ,
( 3 ) D, (A, +
D,( Ak)
Dk( cAk) = c ·D,(~ )
(4)
2. Its value is =
~ ) = (\ (A,) +
on
if the adjacent columns A, and
A.k+t
are equal.
3. Its value is = 1 if all A, are the unit vectors Uk where
1) Henceforth, 0 will denote the zero element
of a field.
1
1
0
0
0
1
;
u,
0
0
0
un
0
1
The question as to whether determinants exist will be left open
for the present. But we derive consequences from the axioms:
a) If we put c :::: 0 in (4) we get: a determinant is 0 if one of
the columns is 0.
b) Dk(Ak)
Dk(Ak +
cA
+ fr a determinant remains unchanged
15 1
if we add a multiple of one column to an adjacent column. Indeed
because of axiom 2.
c) Consider the two columns A, and Ak ... l' We may replace them by
A, and Ak+l + Ak; subtracting the second from the first we may replace
them
by - Ak+l and Ak+l + A, adding the first to the second we now
have .... Ak+l and A, finally, we factor out -1. We conclude: a determinant changes sign if we interchange two adjacent columns.
d) A determinant vanishes if any two of its columns are equal.
Indeed, we may bring the two columns side by side after an interchange
of adjacent columns and then use axiom 2. In the same way as in b)
and c) we may now prove the more general rules:
e) Adding a multiple of one column to another
does
not change
the value of the determinant.
f) Interchanging any two columns changes the sign of D.
14
g) Let ( v1 , v 2 , . , , v n) be a permutation of the subscripts
(1, 2,.... n). If we reanange the columns in D(
Av,I Av,
..
2
. , A, )
n
until they are back in the natural order, we see that
±. D (A" A 2 , ••• , A n ).
D ( Av , Av , ... , A, ) =
1
2
n
Here i is a definite sign that does not depend on the special values
of the A,. If we substitute Uk for A, we see that
D ( U1,
,
1
Uv , ... , U
2
Vn
) =
±.
1 and that the sign depends only on the
permutation of the unit vectors.
Now we replace
tionA~ of
each vector A,
by the following linear combina·
A.,, A2 , ••. , An:
(6) A; = b1 kA 1 + b2 •A, + . . . + bnkAn.
In computing D( A~, A~ , ... , A~) we first apply axiom 1 on A:
breaking up the determinant into a sum; then in each term we do the
same with
A~
and so on. We get
(7) D(A;,A;, ... ,A~)=
v1
L
vl,v2,, ''
I
•v
2
, vn
,.
D(bv 1 Av,bv 2 Av,···'bvnAv)
2
I
,vn
J
1
bv 1·b,
2
• •.•
n
~
2
nn
1
2
where each vi runs independently from 1 to n. Should two of the indices
vi be equal, then D( Av , A, , ... , Av ) = 0; we need therefore keep
1
n
2
only those terras in which (vi, v , . .. , v0
2
)
is a permutation of
( 1, 2, ... , n). This gives
D(A; ,A~, ... ,A~)
(8)
= D( A1, A2
, "
•,
A,).
( v1 ,
~
, , , ,
l/n)
:_ bv t' bv
1
22
. . . . • bv. 0
n
where ( v1 , v2 , ••• , vn) runs through all the permutations of
( 1, 2, ... , n) and where i stands for the sign associated with that
permutation. It is impm1ant to remark that we would have anived at
the
n
·bv D(Av ,A, . . . ,A, )
same formula (8) if our function D satisfied only the first two
n
1.5
of our axioms.
Many conclusions may be derived from (8).
We first assume axiom 3 and specialize the ~ to the unit veetors
of (5). This makes ~I == Bk where Bk is the column vector of
uk
the matrix of the bik' (8) yields now:
(9)
giving us an explicit formula for determinants and showing that they are
uniquely determined by our axioms provided they exist at all.
With expression (9) we return to formula (8) and get
This is the so-called multiplication theorem for determinants. At
the left of (10) we have the determinant of an n-rowed matrix
whose ele-
ments Cik are given by
(11)
c.
tk
=
~" atvbvk'
V=l
cik is obtained by multiplying the elements of the i .. th row of
D(A
1
,
A
2
, ••• ,
An) by those of the k-th column of D(B 1 , 8 2 ,
.•• ,
Bn)
and adding.
Let us now replace D in (8) by a function F(A,
satisfies
only the first two axioms. Comparing with (9) we find
F (A; , A; , ... , A~) ~ F (A 1 ,
... ,
Specializing A, to the unit vectors U k leads to
(12)
, A,) that
F(B 1 ,B 2 ,.
with
,B.)=c·D(B 1 ,B 2 ,.
c = F(U 1 , U 2 ,
... ,
Un).
An )D ( 8 1 , 8 2
, ... ,
B,).
16
Next we specialize (10) in the following way: If
subscript from 1 to n-1 we put A, =
uk for II ,;,
i
is a ce11ain
i, i+ 1
A1 = Ui + Ui+l, A1+1 = 0. Then D( A, A, , .. , A, ) = 0 since one column is
0. Thus,
D( A~ 1 A~,, .. , A~) = 0; but this determinant differs
from that of the elements bJk only in the respect that the i+l-st row
has been made equal to the i-tb. We therefore see:
A determinant vanishes if two adjacent rows are equal.
Each term in (9) is a product where precisely one factor comes
from a given row, say, the i-th. This shows that the determinant is
linear and homogeneous if s;:onsidered as function of this row. If,
finally, we select for eaeh row the corresponding unit vector, th~
terminant is
1 since the matrix is the
=
s~m~
as that in
de-
which the col.,.
umns are unit vectors. This shows that a determinant satisfies oyr
three a:x:ioms if we consider it as function of the row vectors. In view
of the uniqueness it follows:
A determinant remains unchanged if we transpose the row vectors into coltJil'!ll vectors, that is, if we rotate the matrix about its
main diagonal.
A determinant vanishes if
any two rows are equal. It changes
sign if we interchange aqy two rows. It
a multiple of
on~
rerp~ins
unchanged if we add
row to another.
We shall now prove the existence of determinants. For a 1-rowed
mahix a
11
the element a
11
itself is the determinant. Let us assume the
existence of (n • 1) • rowed determinants. If we consider the n-rowed
mahix (l) we may associate with it certain (n • 1) • rowed determinants
in the following way: Let
qik
be a pmticular element in (1). We
17
cancel the i-th row and k-th column in (1) and take the determinant
of the remaining (n .. 1) • rowed matrix. This determinant multiplied by
( -1 )Hk will be called the cofactor of a
ik
and be denoted by Aik.
The distribution of the sign (- l)i+k follows the chessboard pattern,
namely,
+
+ + - +
+ - +
- + - +
' ' '
Let i be any number from 1 to n. We consider the following
function D of the rnauix (1):
D = attAit + ai2Ai2 + · · +
(13)
[t
8 inAin •
is the sum of the products of the i-th row and their cofactors.
Consider this D in its dependence on a given column, say, A,.
.
For v f:- k, A~ depends linearly on A, and a ·v does not depend on it;
for
v
== k, Ark does not depend on A, but a 1k is one element of this
column. Thus, axiom 1 is satisfied. Assume next that two adjacent
columns A, and Ak+ 1 are equal. For 1J ~ k, k + 1 we have then two
equal columns in Ail! so that A, :::: 0. The determinants used in the
computation of A 1
hence, A i
k::::
-Ai
k
and
k+I
Ai
k+
1
whereas a i
holds. For the special case A,
a;v = 0 for
v
t
are the same but the signs are opposite
k ::::
a,
k+r
Thus D :::: 0 and axiom 2
= Uv( v = 1, 2, ..
, n) we have
i while a,, = 1, A; 1 = 1. Hence, D = 1 and
this is axiom 3. This proves both the existence of an n-rowed
18
determinant as well as the truth of formula (13), the so-called development of a determinant according to its i-th row. (13) may be generalized
as follows: In our determinant replace the i-th row by the j-th row and
develop according to this new row. For i
f: j that determinant is 0 and
for i = j it is D:
{
Dforj~i
Oforj,ii
If we interchange the rows and the columns we get the
following fmmula:
Dforho=k
{ Oforh,ik
Now let A represent an n-rowed and B an m-rowed square matrix.
By \ A \, \ B \ we mean their determinants. Let C be a mahix of n rows
and m columns and form the square matrix of n + m rows
(16)
where 0 stands for a zero matrix with m rows and n columns. If we consider the determinant of the matrix (16) as a function ofthecolumns of A
only, it satisfies obviously the first two of our axioms. Because of (12)
its value is c .
A
where c is the determinant of (16) after substituting
unit vectors for the columns of A. This c still depends on B and considered as function of the rows of B satisfies the first two axioms.
Therefore the determinant of (16) is d · A . B
where d is the special
case of the determinant of (16) with unit vectors for the columns of A
as well as of B. Subtracting multiples of the columns of A from
C we can replace C by 0. This shows d = 1 and hence the formula
19
(17)
l
AO
CBI
~
I AI. I Bi
In a similar fashion we could have shown
(18)
A
01
\AI \BI
C B
The formulas (17), (18) are special cases of a general theorem
by Lagrange that can be derived from them. We refer the reader to any
textbook on determinants
since in most applications (17) and (18)
are sufficient.
We now investigate what it means for a matrix if its determinant
is zero. We can easily establish the following facts:
a) If A, A, . , . , An are linearly dependent, then
D (A,. A, ... , A,) = 0. Indeed one of the vectors, say A, is then a
linear combination of the other columns; subtracting this linear combination from the column A, reduces it to 0 and so D = 0.
b) If any vector B can be expressed as linear combination of
A, A, ... , A, then D (A,. A 2 ,
•.• ,
(10) we may select the values for
Al
A,) ~ 0. Returning to (6) and
bik
in such a fashion that every
.~ U1 • For this choice the left side in (10) is 1 and hence
D (A,. A 2 , ••. , A,) on the right side ~ 0.
c) Let A, A, ... , A, be linearly independent and B any other
vector. If we go back to the components in the equation
A 1 X 1 + A 2x 2 + . . . + Anxn+ By= 0 we obtain n linear homogeneous
equations in then + 1 unknowns x 1, x 2 , . . . , xn, y. Consequently,
there is a non-trivial solution. y must be f=- 0 or else the
A 1, A 2 ,
... , ~
would be linearly dependent. But then we can compute
B out of this equation as a linear combination of A, A, . .
An·
20
Combining these results we obtain:
A determinant vanishes if and only if the column vectors (or the
row vectors) are linearly dependent.
Another way of expressing this result is:
The set of n linear homogeneous equations
( i = 1, 2, ... , n)
in n unknowns has a non-trivial solution if and only if the determinant
of the coefficients is zero.
Another result that can be deduced is:
If A I' A 2 ,
... ,
A, are given, then their linear combinations can
represent any other vector B if and only if D (A
, An )
, A,
1
~
o.
Or:
The set of linear equations
(i = 1, 2, ... , n)
has a solution for arbitrary values of the bi if and only if the determinant of aik is
f
0. In that case the solution is unique.
We finally express the solution of (19) by means of determinants
if the determinant D of the aik is
-f
0.
We multiply for a given k the i-th equation with Aik and add the
equations. (15) gives
(20) D· xk = A1kb 1
c
A 2 kb 2
c
Ankbn
(k = 1,2, ... ,n)
and this gives xk. The right side in (12) may also be written as the
determinant obtained from D by replacing the k-th column by
b, b 2'
bn. The rule thus obtained is known as Cramer's rule.
21
II FIELD THEORY
A. Extension Fields.
If E is a field and F a subset of E which, under the operations
of addition and multiplication in E, itself forms a field, that is, if F is
a subfield of E, then we shall call E an extension of F. The relation
of being an extension of F will be briefly designated by F C E. If
a, (3, y, ... are elements of E, then by F(a, (3, y, ... ) we shall mean
the set of elements in E which can be expressed as quotients of polynomials in a,
F (a,
f3,
f3, y, ...
y, .. with coefficients in F. It is clear that
) is a field and is the smallest extension of F which con-
tains the elements a, (3, y, .. We shall call F(a,
fl,
y, . . . ) the field
obtained after the adjunction of the elements a, {3, y, . .. to F, or the
field generated out of F by the elements a, {3, y, .... In the sequel all
fields will be assumed commutative.
If F
c E, then ignoring the operation of multiplication defined
between the elements of E, we may consider E as a vector space over
F. By the degree of E over F, written (E/F), we shall mean the dimension of the vector space E over F. If (E/F) is finite, E will be called
a finite extension.
THEOREM 6. If F, B, E are three fields such that
F
c
B c E, then
(E/F) = (8/F) (E/8) .
Let A1 , A 2 ,. , • , Ar be elements of E which are linearly
independent with respect to Band let C 1, C 2 ,
... ,
C s be elements
22
of B which are independent with respect to F. Then the products Ci Ai
where i = 1, 2, ... , s and j = 1, 2, ... , r are elements of E which are
For if ~ a 11 C 1A; = 0, then
independent with respect to F.
:?(
'·'
2:a 1J C. ) A,. is a linear combination of the A, with coefficients in B
i
l
'
and because the A i were independent with respect to B we have
l:a .. C 1 = 0 for each j. The independence of the C 1 with respect to F
i
lJ
then requires that each ail = 0. Since there are r . s elements CiAi we
have shown that for
each r 2_ ( E/B) and s ::_ ( B/F) the degree ( E/F )
;: r . s. Therefore, ( E/F) ::_ ( 8/F) ( E/8 ). If one of the latter numbers
is infinite, the theorem follows. If both ( E/8) and ( B/F) are finite,
say r and s respectively, we may suppose that the Aj and the C 1 are
generating systems of E and B respectively, and we show that the set
of products Ci A1 is a generating system of E over F. Each A f E can
be expressed linearly in terms of the A. with coefficients in B. Thus,
J
A =
lBJ A .. Moreover, each B. being an element of B can be ex'
J
pressed linearly with coefficients in F in terms of the Ci' i.e.,
Bj = 2:a 1j CFj = 1,2, . . . ,r. Thus, A= ~aij CiAj and the C 1 ~ form
an independent generating system of E over F.
~orollary._lf
F
C
F 1 C F2 C . . . C Fn, then
(Fn/F) c (F/F)·(F2 /F,) . . . (Fn/Fn_ 1 ).
B.
Polvnomials.
An expression of the form a 0 xn +a 1 xn-t+ ... + a 0 is called a
polynomia~
in F of degree n if the coefficients
23
a
0
, ... ,
a,., are elements of the field F and a
0
f: 0. Multiplication and
addition of polynomials are performed in the usual way 0 .
A polynomial in F is called reducible in F if it is equal to the
product of two polynomials in F each of degree at least one. Polynomials which are not reducible in F are called ineducible in F.
If f (x ) = g(x) . h (x ) is a relation which holds between the
polynomials f (x ), g (x ), h (x ) in a field F, then we shall say that
g (x ) divides f (x ) in F, or that g ( x ) is a factor off ( x ). It is readily
seen that the degree of f(x) is equal to the sum of the degrees of
g (x ) and h (x ), so that if neither g ( x ) nor h ( x ) is a constant then
each has a degree less than f(x). It follows from this that by a finite
number of factorizations a polynomial can always be expressed as a
product of ineducible polynomials in a field F.
For any two polynomials f (x ) and g (x ) the division algorithm
holds, i.e., f(x) = q(x)·g(x) + r(x) where q(x) and r(x) are
unique polynomials in F and the degree of r (x ) is
less than that of
g(x). This may be shown by the same argument as the reader met in
elementary algebra in the case of the field of real or complex numbers.
We also see that r(x) is the uniquely determined polynomial of a degree less than that of g (x ) such that f(x) - r (x ) is divisible by
g (x ). We shall call r (x ) the remainder off (x ).
1) If we speak of the set of all polynomials
of degree lower than n, we shall agree to
include the polynomial 0 in this set,
though it has no degree in the proper sense.
24
Also, in the usual way, it may be shown that if a is a root of
the polynomial f (x ) in F than x -
a is
a factor off (x ), and as a con-
sequence of this that a polynomial in a field cannot have more roots
in the field than its degree.
Lemm<!_. If f(x) is an irreducible polynomial of degree n in F,
then ther~ do not exist two polynomials each of degree less than n in
F whose _product is divisible by f(x).
Let us suppose to the contrary that g(x) and h(x) are polynomials of degree less than n whose product is divisible by f(x).
Among all polynomials occmring in such pairs we may suppose g(x)
has the smallest degree. Then since f(x) is a factor of g(x) . h (x )
there is a polynomial k(x) such that
k(x)·f(x) = g(x)·h(x)
By the division algorithm,
f(x) = q(x).g(x) + r(x)
where the degree of r (x ) is less than that of g(x) and r (x )
~
0
since f(x) was assumed ineducible. Multiplying
f(x) = q(x).g(x) + r(x)
by h (x ) and transposing, we have
r (X) · h (X) = f (X)· h (X )-q (X ) · g (X) · h (X) - f (X)· h (X )-q (X )· k (X) ·f (X)
from which it follows that r(x) . h (x ) is divisible by f (x ). Since r (x )
has a smaller degree than g(x), this last is in contradiction to the
choice of g (x
)~
from which the lemma follows.
As we saw, many of the theorems of elementary algebra
hold in any field F. However, the so-called Fundamental
Theorem of Algebra, at least in its customary form, does not
hold. It will be replaced by a theorem due to Kronecker
25
which guarantees for a given polynomial in F the existence of an extension field in which the polynomial has a root. We shall also show
that, in a given field, a polynomial can not only be factored into irredllcibl~ factors, but that this
factorization
is unique up to a constant
factor. The uniqueness depends on the theorem of Kronecker.
C. Algebraic Elements.
Let F be a field and E an extension field of F. If a is an element of E we may ask whether there are polynomials with coefficients
in F which have a as root. a is
tkere are
~alled
algebraic with respect to F if
such polynomials. Now let a be algebraic and select among all
polynomials in F which have
a
as root
one, f(x), of lowest degree.
We mpy assume that the highest coefficient of f(x) i~ 1. We con-
tend that this f(x) ~~ uniquely determined, that it is irreducible and
that each polynomial in F with the r1>0t a is divisible by f c~
). If. in-
deed, g ( ~ ) i& a pQ!ynomial in F with g(a) = 0, we may divide
g(x) = f(x)q(x) t r(x) where r(x) ha~ a degree sm~ller than that
of f(x). Substituting x = a we get r ( n l
i<lenti~ally
0 since otherwis@ r ( ~ )
lo\Ver degree thap f (x ),
the uniqueness of
f (x
So g ( x
)
woul~
j~
=
0. NPW r(x) has to be
h§Ve the root
divisible by
f ( • ),
a
and be of
Th!~
also shows
), Iff ( N ) were not irreducible, one of the factors
Would have to vanish for x =a contraqict\ng again the choice off ( y ),
We consider now the subset Eo of the following
0 ofE:
element~
26
(J = g(a) =
Co+ Cla + c2a2
+ ... + cn-Ian-1
where g(x) is a polynomial in F of degree less than n (n being the degree of f(x)). This set
tf, is closed under addition and multiplication.
The latter may be verified as follows:
If g (x ) and h (x ) are two polynomials of degree less than n we
put g(x)h(x) = q(x)f(x) + r(x) and
hence g(a)h(a) = r(a).
Finally we see that the constants c , c , .. , c
are uniquely deter1
'
n- 1
mined by the element 0. Indeed two expressions for the same f) would
lead after subtracting to an equation for a of lower degree than n.
We remark that the internal structure of the set E 0 does not de-
pend on the nature of a but only on the ineducible f (x ). The knowledge
of this polynomial enables us to perform the operations of addition and
multiplication in our set E 0 • We shall see very soon that E 0 is a field;
in fact, Eo is nothing but the field F(a). As soon as this is shown we
have at once the degree, ( F (a) /F), determined as n, since the space
F(a) is generated by the linearly independent 1, a,
a2, . .. ,
an-I,
We shall now try to imitate the set E 0 without having an extension field E and an element a at our disposal. We shall assume only
an ineducible polynomial
as given.
We select a symbol
I; and let E 1 be the set of all formal
polynomials
g(f;) =co+ elf;+ .. + cn_,f;n-1
of a degree lower than n. This set forms a group under
addition. We now introduce besides the ordinary multiplication
27
a new kind of multiplication of two elements g (,f) and h (,f) of E 1
denoted by g (,f) x h
(<fl.
It is defined as the remainder r (.f) of the
ordinary product g (,f) h (,f) under division by f (f). We first remark
that any product of m terms g 1 ( ,f), g 2 ( f),, .. 1 gm(
mainder of the ordinary product g 1( ~) g ,(
f) is again there-
0 ... g'"( ~).
This is true by
definition for m = 2 and follows for every m by induction if we just
prove the easy lemma: The remainder of the product of two remainders
(of two polynomials) is the remainder of the product of these two
polynomials. This fact shows that our new product is associative and
commutative and also that the new product g
(
1
f)
x g 2( ,f) x , , , x g
mC<f)
will coincide with the old product g 1( ,f) g 2( ~) •. , gm( ,f) if the latter
does not exceed n in degree. The distributive law for our multiplication
is readily verified.
The set E 1 contains our field F and our multiplication in E1 has
for F the meaning of the old multiplication. One of the polynomials of
E 1 is
f
Multiplying it i-times with itself, clearly will just lead to
as long as i
< n.
<f'
For i : : : n this is not any more the case since it
leads to the remainder of the polynomial
I;.
This remainder is
= - a n-
t
n-1
- an- ~n-2 - ' ' ' - ao.
We now give up our old multiplication altogether and keep only
the new one; we also change notation, using the point (or juxtaposition)
as symbol for the new multiplication.
Computing in this sense
Co
+
c
CtS
+
/:2
C2t.:,
+'
·+
en-!
Cn-tS
will readily lead to this element, since all the degrees
28
involved are below n. But
Transposing we see that f (
el
= 0.
We thus have constructed a set E
1
and an addition and multipli-
cation in E 1 that already satisfies most of the field axioms. E 1 contains
F as subfield and
satisfies the equation f ( ~) = 0. We next have to
e
show:
If g (
f) i 0 and h (
~)are given elements of E
1,
there is
an element
X( f)=
•. + x 1 ~
x._,f•·'
+ ... +
in E l such that
&<e).
xcel =
h(eJ.
To prove it we consider the coefficients x. of X ( ~) as unknowns and
'
.
compute nevertheless the product on the left side, always reducing
higher powers of ~ to lower ones. The result is an expression
L
o
+ L 1 ~ + . . + Ln-t
en-t where
each L 1 is a linear combination of
of the Xr with coefficients in F. This expression is to be equal to
h ( {); this leads to the n equations with n unknowns:
Lo = bo, Ll = bl' · · ·
1
Ln-l = bn-1
where the b1 are the coefficients of h (f). This system will be soluble
if the conesponding homogeneous equations
L. = 0, L, = 0, ' ..
have
I
L •. ,
0
only the trivial solution.
The homogeneous probleiP would occur if we should ask for
the set of elements X(Q) satisfying g (
0 . X ( el
= 0. Going back
for a moment to the old multiplication this would mean that the
ordinary product g (
0
X ( eJ has the remainder 0, and is
29
therefore divisible by f(t). According to the lemma, page 24, this is
only possible for X (f) = 0.
Therefore E 1 is a field.
Assume now that we have also our old extension E with a root
a of f(x), leading to the set E,. We see that E 0 has in a certain sense
the same structure as E
1
if we map the element g (,;)of E
1
onto the
element g(a) of E0 • This mapping will have the prope11y that the image
of a sum of elements is the sum of the images, and the image of a
product is the product of the images.
Let us therefore define: A mapping a of one field onto another
which is one to one in both directions such that
a(a+f3) = a(a) + a({3) and a(a·{3)= a(a)· a({3) is called an
isomorphism. If the fields in question are not distinct - i.e., are both
the same field -
the isomorphism is called an automorphism. Two
fields for which there exists an isomorphism mapping one on another
are called isomorphic. If not every element of the image field is the image
under a of an element in the first field, then o is called an isomorphism
of the first field into the second. Under each isomorphism it is clear
that a(O) = 0 and a( 1) = 1.
We see that Eo is also a field and that it is isomorphic to E,.
We now mention a few theorems that follow from our discussion:
THEOREM 7. (Kronecker). If f (x ) is a polynomial in a field F,
there exists an extension E of F in which f(x) has a root.
30
Proof: Construct an extension field in which an irreducible
factor off ( x ) has a root.
THEOREM 8. Let
a be
an isomorphism mapping a field F on a
field F' Let f (x ) be an irreducible polynomial in F and f
responding polynomial in F
1 •
If E = F (
f3) and E
a can
= F
1
(
= 0 in E and
(x ) the cor-
f~') are exten-
f
1
be extended to an isomorphism between E and E
1
sions ofF and F', respectively, where f(f3)
then
1
1
(
f3 ')
= 0 in E'
Proof: E and E' are both isomorphic to E 0 •
D.
Splitting
Fields.
If F, B and E are three fields such that F C B C E, then we
shall refer to B as an intermediate field.
If E is an extension of a field F in which a polynomial p(x) in F
can be factored into linear factors, and if p(x) can not be so factored
in
any
intermediate field, then we call E a splitting field for p(x). Thus,
if E is a splitting field of p(x), the roots of p(x) generate E.
A splitting field is of finite degree since it is constructed by a
finite number of adjunctions of algebraic elements, each defining an
extension field of finite degree. Because of the corollary on page 22,
the total degree is finite.
THEOREM 9. If p(x) is a polynomial in a field F, there exists
a splitting field E of p(x).
We factor p (x ) in F into irreducible factors
f,(x)
f,( x)
= p(x). If each of these is of the first
degree then F itself is the required splitting field. Suppose
then that f 1 ( x) is of degree higher than the first. By
31
Theorem 7 there is an extension F 1 ofF in which f
Factor each of the factors f
1
(
(
1
x ) has a root.
x), ... , fr ( x ) into irreducible factors in
F 1 and proceed as before. We finally arrive at a field in which p (x)
can be split into linear factors. The field generated out of F by the
roots of p(x) is the required splitting field.
The following theorem asserts that up to isomorphisms, the
splitting field of a polynomial is unique.
THEOREM 10. Let
the field
F' ,
~~ith
a
be an isomorphism mapping the field F on
Let p (x ) be a polynomial in F and p
1
(x ) the polynomial
coefficients corresponding to those of p (x ) under
let E be a splitting field of p(x) and E
Under these conditions the isomorphism
isomorphism between E and E
If f(x) is an irreducible
1
1
1) (
a can
in which f(a) = 0. Then (a-a
be extended to an
•
factor of p(x) in F, then E contains a
x-a 2 ).
f(x) as a polynomial in E and
Finally,
a splitting field of p 1 (x).
root of f( x ). For let p (x )=(x-a ~ (x-a, )
p(x) in E. Then ( x-a
a.
•(
(x-a ,) be the splitting of
x-a ,) = f(x) g(x). We consider
construct the extension field B = E(a)
1 )·
(a-a
2 )· .•.
·(a-a,)= f(a)· g(a) = 0
and a-a1 being elements of the field B can have a product equal to 0
only if for one of the factors, say the first, we have
a-a 1 = 0. Thus,
a = a 1 , and a 1 is aroot of f(x).
Now in case all roots of p(x) are in F, then E = F and p(x)
can be split in F. This factored form has an image in F
splitting, of p 1 (x), since the isomorphism
a
I
which is a
preserves all operations
of addition and multiplication in the process of multiplying out the
factors of p(x) and collecting to get the original form. Since p ' (x)
can be split in F' , we must have F
1
::::
E
1
•
In this case,
a itself is
the required extension and the theorem is proved if all roots of p(x)
ar~
in F.
We proceed by complete induction. Let us suppose the theorem
proved for all cases in which the number of roots of p(x) outside of F
is less than n
> 1, and suppose that p (x ) is a polynomial having n
roots outside of F. We factor p (x ) into ineducible factors in F;
p(x) = f,(x) f (x) .. f,(x). Not all of these factors can be of
2
degree 1, since otherwise p (x ) would split in F, contrary to assumption. Hence, we may suppose the degree of f
1(
x) to be r > 1. Let
f\(X).f'2(x) . . . fO:,(x) = p'(x) be the factorization of p'(x) into
the polynomials comespondng to f
is ineducible in F
1 ,
(
1
x ) , ... , f m ( x ) under a.
for a factorization of
f;
(x) in F
1
£; (x )
would induce
1)
under a-1 a factorization of f,(x), which was however taken to
be irreducible.
By Theorem 8, the isomorphism
a can
be extended to an isomor-
phism a 1, between the fields F (a) and F ' (a 1 ).
Since F C F(a), p(x) is a polynomial in F (a) and E is a
splitting field for p(x) in F(a). Similarly for p ' (x). There are now
less than n roots of p (x ) outside the new ground field F (a). Hence
by our inductive assumption a 1 can be extended from an isomorphism
between F (a) and F 1 (a ' ) to an isomorphism a 2 between E and E
Since
a1
is an extension of
a,
and
a2
an extension of
a 2 is an extension of a and the theorem follows.
t)
See page 38 for the definition of
a
a 1, we
1
•
conclude
33
Corollary. If p ( x) is a polynomial in a field F, then any two
splitting fields for p ( x ) are isomorphic.
This follows from Theorem 10 if we take F = F ' and a to be the
identity mapping, i.e., a(x) = x.
As a consequence of this corollary we see that we are justified
in using the expression "the splitting field of p ( x )" since any two
differ only by an isomorphism. Thus, if p (x ) has repeated roots in one
splitting field, so also in any other splitting field it will have repeated
roots. The statement up ( x) has repeated roots" will be significant
without reference to a particular splitting field.
E. Unique Decomposition of Polynomials into Irreducible Factors.
THEOREM
11. If p( x) is a polynomial in a field F, and if
p(x) = p 1(x)·p 2 (x)· .. , ·p,(x) = q 1(x)·q 2(x)· .... q (x) are two
factorizations of p ( x) into irreducible polynomials each of de~ree at
least one, then r = s and after a suitable change in the order in which
the q's are written, p,(x)
~
c,q.(x), i = 1,2,., ,r, and c,' F.
Let F (a) be an extension of F in which p
1(
a) ~ 0. We may
suppose the leading coefficients of the p,( x) and the q1(X) to be 1, for,
by factoring out all leading coefficients and combining, the constant
multipliE~r
on each side of the equation must be the leading coefficient
of p ( x ) and hence can be divided out of both sides of the equation.
Since 0 = p 1(a)· p 2 ( a) · , , , · P, (a) = p (a)
~
q 1( a) · . , , · q ,(a) and
since a product of elements of F (a) can be 0 only if one of these is 0,
it follows that one of the q,( a), say q 1( a), is 0. This gives (see page
25) p 1(>) = q 1(x ). Thus p 1(x). p 2(x) ..... p,(x)
= p 1(x)·q 2 (x)·,, ·q,(x) or
H
p 1(x) ·[p 2(x) · ..
·p,(x)- q 2 (x) · ..
·q,(x)] =
0.
Since the product
of two polynomials is 0 only if one of the two is the 0 polynomial, it
follows that the polynomial within the brackets is 0 so that
· Pr ( x) = q 2( X)· ... · q 5( X). If we repeat the above argument
p,(x)
r times we obtain pi ( x) = q i (X). i =
1, 2,. .
r. Since the remaining
q's must have a product 1, it follows that r-: s.
F.
~roup
C~l!_racter~.
If G is a multiplicative group, F a field and
a
a homomorphism
mapping G into F, then a is called a character of G in F. By homomorphism is meant a mapping
a such
that for a,
f3
any two elements of G,
a( a )·a({3) = a(a·{J) and a(a) ~ Oforanya.
(If o(a) = 0 for one element a, then o(x) = 0 for each x ( G, since
a( ay) = a( a:)· o(y) ""0 and ay takes
all
values in G when
y
assumes
all values in G).
The characters a
elements a
1
,
a, .
1
,
a 2 ,.
.
,
a nare called dependent if there exist
. , a, not all zero in F such that
. + ana n( x) :::
0 for each x ( G. Such a de-
pendence relation is called non-trivial. If the characters are not
dependent they are called independent.
THEOREM 12. If G is a group and a 1' a 2, •.. , an are n mutually distinct characters of G in a field F, then a 11 a 2,. .
an
are independent.
One character cannot be dependent, since a
a1
1a 1(
x) ::: 0 implies
0 due to the assumption that a (x) ~ 0. Suppose n
1
> 1.
35
We make the inductive assumption that no set of less than n distinct
characters is dependent. Suppose now that
a 0 a 0 ( x) :::: 0 is a non-trivial dependence
t
between the a's.
None
of the elements ai
is
zero, else we should have
a dependence between less than n characters contrary to our induetive assumption. Since o
a in Gi such
that
a1 (a)
~
1
and an are distinct, there exists an element
an( a). Multiply the relation between the
a's b
y n a ·We obtain a relation
(*)
b 1 a 1 (x)t
+b0 _ 1 u 0 _ 1 (x)to,(x)
Replace in this relation x by ax. We have
0,
Subtracting the latter from ( *) we have
(**) [b 1
an Car'b 1 a 1 (a)]a 1 (x)'.
+ C 0 _ 1 0 0 _ 1 (x)
o .
=
The coefficient of a 1 (x ) in this relation is not 0, otherwise we should
have b 1 = o-0 (a)" 1 b 1 a 1 (a), so that
an (a)b 1 = b 1 a 1 (a)= u 1 (a)b 1
and since b 1 ~ 0, we get a,( a) =
a 1 (a) contrary to the choice of a.
Thus, (* *) is a non-trivial dependence between
a
l'
a2,
,
•
an-
1 which
is contrary to our inductive assumption.
Corollary. If E and E
1
are two fields, and a 1 , a2
mutually distinct isomorphisms mapping E into E
1 ,
an are n
, .. ,
then u 1
,
,
an
are independent. (Where ttindependent" again means there exists no
non-trivial dependence a
every x
f
1a 1
(x ) + . + anan (x ) == 0 which holds for
E).
This follows from Theorem 12, since E without the 0 is a group
36
a's defined in this group are mutually distinct characters.
and the
If a 1 , o2 1
then
each
••• ,
an
are isomorphisms of a field E into a field E'
element a of E such that
is called a fixed point of E under
a 1 (a) = o,(a) =
a1 ,
on . This name is
02 , .
chosen because in the case where the a's are automorphisms and a1
is the identity,
i.e. 1 a1 (x)
= x, we have
ai
(x) = x for a
fixed point.
Lemma. The set of fixed points of E is a subfield of E. We
shall call this subfield the fixed field.
For if a and b are fixed points, then
a1 (a +b)= a 1 (a) + a1 (b) =";(a)+ Oj (b)="; (a+ b) and
a,(a·b) = a 1 (a)·a,(b) = "; (a)·a;(b) =a; (a·b).
Finally from
=
",ca-
1
)
a 1 (a)
=
a,
(a) we have (a,(a))- 1 = (a 1 (alr'
=a; (a-').
Thus, the sum and product of two fixed points is a fixed point, and
the inverse of a fixed point is a fixed point. Clearly, the negative of a
fixed point is a fixed point.
THEOREM 13. If a 1'.
of a field E into a field E
1
,
.
1
an are n mutually distinct isomorphisms
and if F is the fixed field of E, then
(E/F) >_n.
Suppose to the contrary that ( E/F) = r < n. We shall show that
we are led to a contradiction. Let
w 1,
w 21 . . . , wr be a generating sys-
tern of E over F. In the homogeneous linear equations
37
u 1 (w 1 )x 1 + a 2 (w 1 )x 2 + ..
+ a 0 (w 1 )xn = 0
u 1 (w 2 )x 1 + o- 2 (w 2 )x 2 + ..
+ an(w,)xn = 0
a 1 (cur)x 1 + a 2 (wr)x 2 + ... + a 0
(W 1 )X
= 0
0
there are more unknowns than equations so that there exists a nontrivial solution which, we may suppose, x
any element a in Ewe can find a 1 , a 2 ,
1,
x
•.. ,
denotes. For
2 , . . . , X0
a, in F such that
a ::. a 1t:v 1 + ... + a/ur. We multiply the first equation by
the second by a 1 (a
2
),
and so on. Using that a 1
a 1 (aJ) = a 1 (a 1 )and also that a (a)
1
we obtain
a 1 (arwr)x
1
+ ... +
0
0
a;(w 1)
f.
F,
a 1 (a
1
),
hence that
= a (a 1w 1 ),
1
(arwr)x0 = 0.
Adding these last equations and using
ai(a 1 U)
1)
1-
a 1 (a 2 (t.~ 2 ) + ... +
aJarcu) =
a 1 (a 1w 1 + ... + arwr) = ai(a)
we obtain
a 1 (a)x 1 + a 2 (a)x 2 + . . . + un(a)xn = 0.
This, however, is a non-trivial dependence relation between a 11 a 2 , .
which cannot exist according to the corollary of Theorem 12.
Corollary. If a
1,
a 2,
••• ,
an are automorphisms of the field E, and
F is the fixed field, then ( E/F) > n.
If F is a subfield of the field E, and
shall say that
a
leaves F fixed if for
a
an automorphism of E, we
each element a of F, o(a) = a.
a0
38
If a and r are two automorphisms of E, then the mapping o-( r( x)) written
briefly ar is an automorphism, as the reader may readily verify.
[E.g., <n(x y) ~ a(r(x · y))
We shall call
(a(x)
~
= a(r(x)· r(y)) = a(r (x)) · a(r(y))].
ar the product of a and r. If a is an automorphism
y), then we shall call a· 1 the mapping of y into x, i.e.,a· 1(y)
~
the inverse of a. The reader may readily verify that a·' is an automor·
phism. The automorphism I (x)
~
x shall be called the
unit automorphism.
Le~.
If E is an extension field of F, the set G of automorphisms
which leave F fixed is a group.
The product of two automorphisms which leave F fixed clearly
leaves F fixed. Also, the inverse of any automorphism in G is in G.
The reader will observe that G, the set of automorphisms which
leave F fixed, does not necessarily have F as its fixed field. It may be
that certain elements in E which do not belong to F are left fixed by
every automorphism which leaves F fixed. Thus, the fixed field of G
may be larger than F.
G. Applications and
Examples to Theorem 13.
Theorem 13 is very powerful as the following examples show:
1) Let k be a field and consider the field E = k (x ) of all
rational functions of the variable x. If we map each of the functions
f (x) of E onto f(!) we obviously obtain an automorphism of E. Let US
X
consider the following six automorphisms where f (x ) is mapped onto
1
1 ) and f
f (x) (identity), f ( 1-x), f ( ·), f (1·! ), f CX
X
1-X
C-"-l and call
X-1
F the
x
39
fixed point field. F consists of all rational functions satisfYing
(1)
It
1 = f(l-x)
1 = f(r:xl
1 = f( x-1
X )
f(x) = f(l-x) = f(x)
.
suffices
to check the first two equalities, the others being conse-
quences. The function
I= I(x)- (x'- X+ 1 )'
x 2 (x-1) 2
belongs to F as is readily seen. Hence, the field S = k (1) of all
(2)
rational functions of 1 will belong to F.
We contend: F = Sand ( E/F) = 6.
lndeed, from Theorem 13 we obtain (E/F) 2 6. Since S C Fit
suffices to prove (Ef S) < 6. Now E
= Sx). It is thus sufficient
to
find some 6-th degree equation with coefficients in S satisfied by x.
The following one is obviously satisfied;
(x 2 -x+1) 3 -1-x 2 (x-1) 2 = 0.
The reader will find the study of these fields a profitable exer-
cise. At a later occasion he will be able to derive all intermediate fields.
2) Let
k be a field and E
= k(x 1 ,x 2 ,
rational functions of n variables x 1 , x 2 ,
... , X
...
•
0
,xn)the field of all
If ( v 1 , v 2 ,
permutation of { 1, 2, ... , n) we replace in each function f (x
of E the variable
X1
by
X
, X
vl
2
by X
~
,.
.
... ,
1
,
vn ) is a
x 2 , . . . , xn)
, X by XV . The mapping of E
n
n
onto itself obtained in this way is obviously an automorphism and we
may construct n ! automo:rphisms in this fashion (including the identity).
Let F be the fixed point field, that is, the set of all so-called
"symmetric functions." Theorem 13 shows that ( E/F) >_ n ! . Let us introduce the polynomial:
(3) f(t) = (t-x 1 )(t-x
2
) ...
(t-x,)
tn
+ a
1 t n-1
+ . +
an
40
wherea 1 = ... (x 1 + x 2 + ... + x,); a 2
(x 1 x 2 + x 1 x3 + ... +
:: +
xn-lxn)
and more generally ai is ( • 1)' times the sum of all products of i differerentvariables of the set x 1 , x 2 ,
... ,
xn.Thefunctionsa, a 2 ,
... ,
an
are called the elementary symmetric functions and the field
S = k ( a1 , a2 ,
.•• ,
an ) of all rational functions of a,, a 2 ,
an is
... ,
obviously a part of F. Should we suceed in proving ( Ef S )
< n ! we
would have shown S = F and ( E/F) = n ! .
We construct to this effect the following tower of fields:
S = Sn C Sn-t C S 0 _ 2
...
C S 2 C S1 = E
by the definition
(4)
sn = S; s± = S(xl+t ,xi+2····,xn) = s±•t (xt+t ).
It would be sufficient to prove ( S 1 _1 /S± ) <_ i or that the generator x 1
for Si-l out of ~ satisfies an equation of degree i with coefficients
inS1 .
Such an equation is easily constructed. Put
f(t)
FH! (t)
(5) F, ( t) =
(t-xl+l )(t-x,+2 ) ... (t-xn) = (t-x,+l)
and F n
(
t) = f ( t ). Perfonning the division we see that F 1 (t ) is a
polynomial in t of degree i whose highest coefficient is 1 and whose
coefficients are polynomials in the variables
a1 , a2,
... ,
a,, and
xi+l,
xi+2
in these expressions. Now
Now let g(x 1 , x 2 ,
Since F1 (
'S ) =
, . . . , xn.
Only integers enter as coefficients
xi is obviously a root of F1 (t )
••• , xn)
be a polynomial in x 1 , x 2 ,
0 is of first degree in
'S , we
polynomial of the~ and of x 2 , x 3 ,
. . . , xn.
in g( x 1 , x 2 ,
) =
••• , xn ).
Since F2 ( x 2
can express
= 0.
•.. , xn.
'S
as a
We introduce this expression
0 we can express x~ or higher
41
powers as polynomials in x 3 ,
... ,
xn and the a 1 • Since F 3
x3) = 0
(
we can express xj and higher powers as polynomials of x 4 , x 5 ,, .. , xn
and the ai. Introducing these expressions in g( x 1 , x 2 , •.. , xn) we see
that we can express it as a polynomial in the
the degree in xi is below i. SJ g( x 1 , x 2 ,
~
... , X
and the
)
0
~
such that
is a linear combination
of the following n ! terms:
vn
(6)
... X
0
where each vi :S_ i - 1.
The coefficients of these terms are polynomials in the ai . Since the
expressions (6) are linearly independent in S (this is our previous
result), the expression is unique.
This is a generalization of the theorem of symmetric ftu1ctions in
its usual form. The latter says that a symmetric polynomial can be
written. as a polynomial in a 1 , a, ... , a,. Indeed, if g(x 1 ,
...
,x 0
)
is
symmetric we have already an expression as linear combination of the
terms (6) where only the term 1 conesponding to v1 =
has a coefficient
-f 0 inS, namely, g( x 1 ,
is a polynomial in a, a 2 ,
•.. ,
... ,
l.-'
2
= ... = vn = 0
xn ). So g( x 1 , x 2 , •.. , xn)
a,.
Hut our theorem gives an expression of any polynomial, symmetric
or not.
H. Nmmal Extensions.
An extension field E of a field F is called a normal extension if
the group G of automorphisms of E which leave F fixed has F for its
fixed field, and ( E/F) is finite.
Although the result in Theorem 13 cannot be sharpened in general,
42
there is one case in which the equality sign will always occur, namely,
in the case in which
a1 , a2 , . . .
, an is a set of automo:rphisms which
form a group. We prove
THEOREM 14. If a 1 , u2 , •. . ,
an is a group of automo:rphisms of a
_field E ru:d ifF is the fixed field of u 1
Ifa 1 ,a 2 ,
•••
,a, ... , a., then (E/F)
= n.
,a0 is a group, then the identity occurs, say, (]
1
=I.
The fixed field consists of those elements x which are not moved by
any of the a's, i.e., a i ( x) = x, i ::::: 1, 2, ... n. Suppose that ( E/F ) > n.
Then there exist n + 1 elements a 1 , a 2 ,
. . . , an+l
of E which are
linearly independent with respect to F. By Theorem 1, there exists a
non-trivial solution in E to the system of equations
0
xl al (al ) + x2 al (a2 ) + ' , ' + xn+l at (an+l ) -::
Xl a2
(a I
)
+
x2
a2 ( a2
+ ... +
)
0
xn+I a2 ( an+l ) =
(' )
x1
a. ( a 1 )
+ x 2 an ( a2
)
+ , , , + xn+l "" (an+! ) = 0
We note that the solution cannot lie in F, otherwise, since a is the
1
identity, the first equation would be a dependence between a,,
Among all non-trivial solutions x 1 x
'
2'
... , x
n+l
.
.
.
,
a
n+l
we choose one
which has the least number of elements different from 0. We may suppose this solution to be a 1 , a 2 ,
... ,
a, 0, ... , 0, where the first r
terms are different from 0. Moreover, r
implies a 1 = 0 since
a 1 (a,
)
f.
1 because a, a 1 (a 1 ) = 0
= a 1 -/:. 0. Also, we may suppose ar = 1,
since if we multiply the given solution by a; 1 we obtain a new solution
in which the r- th term is 1. Thus, we have
43
(*) a 1a,(a 1 ) + a2 a1(a 2 ) + .. , + a,_ 1a,(a,_ 1 ) + a1 (a,)
fori=
1, 2, ... , n.
these, say a
1
,
Since a 1 ,
. . . , ar-l
~
0
cannot all belong to F, one of
is in E but not in F. There is an automorphism ak for
which ak( a,) f. a 1 . If we use the fact that a 1 , a 2 ,
. .. ,
an form a group,
we see 17k·a 1 ,ak·a 2, ... ,ak·anisapermutationofa 1 ,u 2, . . . ,an.
Applyjng ak to the expressions in ( *) we obtain
ak(a 1 ).akaj(a 1 )
+ ,,, +
ok(ar_ 1 )·akaj(ar-I)
for j = l, 2, , .. , n, so that from
and if we subtract (
akaj
+
akaj(ar)
= 0
=a1
* * ) from ( * ) we have
[a 1 - ak(a 1 )l. aJa 1
)
+ ...
+ la,_ 1 - ak(a,_ 1 )]a,(a,_1 ): 0
which is a non- trivial solution to the system
C ) having
fewer than
elements different from 0, contrary to the choice of r.
Corollary 1. If F is the fixed field for the finite group G, then
each automorphism a that leaves F fixed must belong to G.
( E/F) = order of G = n. Assume there is a
a not
in G. Then F
would remain fixed under the n + 1 elements consisting of (] and the
elements of G, thus contradicting the corollary to Theorem 13.
Corollary 2. There are no two finite groups G1 and G 2 with the
same fixed field.
This follows immediately from Corollary 1.
If f(x) is a polynomial in F, then f(x) is called separable if its
irreducible factors do not have repeated roots. If E is an extension of
44
the field F, the element a of E is called separable if it is root of a
separable polynomial f(x) in F, and E is called a separable extension
if each element of E is separable.
THEOREM 15. E is a normal extension of F if and only if E is
!he spli!ting field of a separable polynomial p(x) in F.
Sufficiency. Under the assumption that E splits p (x) we prove
that E is a normal extension ofF.
If all roots of p(x) are in F, then our proposition is trivial, since
then E = F and only the unit automorphism leaves F fixed.
Let us suppose p(x) has n > 1 roots in E but not in F. We make
the inductive assumption that for all pairs of fields with fewer than n
roots of p(x) outside of F our proposition holds.
Let p(x) = p 1 ( x). p 2 ( x). . . . . p,( x) be a factorization of p(x)
into irreducible factors. We may suppose one of these to have a degree
greater than one, for otherwise p(x) would split in F. Suppose deg
p,(x) = s > 1. Let a be a root of p,(x). Then (F(a, )/F) = deg p,(x)
1
If we consider F (a
1
)
as the new ground field, fewer roots of p ( x) than
n are outside. From the fact that p(x) lies in F(a, ) and E is a splitting field of p(x) over F( a 1
),
it follows by our inductive assumption
that E is a normal extension of
F (a,
).
Thus, each element in E which
is not in F(a, ) is moved by at least one automorphism which leaves
F( a, ) fixed.
p (x) being separable, the roots a , a 2 , • • • , as of p (x) are a
1
1
distinct elements of E. By Theorem 8 there exist isomorphisms
s.
45
au a 2 ,
•. , ,
a, mapping F ( a 1 ) on F ( a 1 ), F ( a 2 ),
••• ,
F (a,),
respectively, which are each the identity on F and map u 1 on
a
1
,
a , ••• , as respectively. We now apply Theorem 10. E is a splitting
2
field of p(x) in F(a, ) and is also a splitting field of p(x) in F(a, ),
Hence, the isomorphism ai, which makes p( x ) in F ( a ) correspond to
1
the same p(x) in F( a., ), can be extended to an isomorphic mapping of
E onto E, that is, to an automorphism of E that we denote again by ai.
Hence, a
1
,
a
2
,. •
,
as are automorphisms of E that leave F fixed and
map a 1 ontO al'a , •.. a,.
2
Now let
8 be
an element that remains fixed under all automor·
phisms of E that leave F fixed. We know already that it is in F (a
1
)
and hence has the form
e
=
co
+
c1a 1
-t
c, a,2 -l . . .
+
c s-1 a 1s-1
where the ci are in F. If we apply a 1 to this equation we get, since
a,(O)=O:
The polynomial c
s- 1
xs-l
+ c s-2 x
has therefore the s distinct roots a 1 , a
2
s-2
,.
+
+ c x
1
,
+ (c
0
as. These are more than
its degree. SJ all coefficients of it must vanish, among them c
This shows
8 in
e)
0
-
e.
F.
Necessity. If E is a normal extension of F, then E is spiitting
field of a separable polynomial p(x). We first prove the
Lemma. If E is a normal extension of F, then E is a separable
extension of F. Moreover any element of Eisa root of an equation over
F which splits completely in E.
46
Let a 1 , a 2 ,
. . . , a~
be the group G of automo:rphisms of E whose
fixed field is F. Let a be an elementofE, and let a, a 2 ,a 3 ,
the set of distinct elements in the sequence a
1
(
...
,ar be
a), a 2 ( a), ... , a 5 ( a).
Since G is a group,
a;(a,) = a;<a.(a)) = a;ak(a) = a,(u) = a,.
Therefore, the elements a,a 2 ,
... ,
ar are permuted by the automo:rphisms
of G. The coefficients of the polynomial f(x) = ( x-a) ( x-u
2) • . .
(
x-a,)
are left fixed by each automorphism of G, since in its factored form the
factors of :flx) are only permuted. Since the only elements of E which
are left fixed by all the automorphisms of G belong to F, f(x) is a
polynomial in F. If g(x) is a polynomial in F which also has a as root,
then applying the automorphisms of G to the expression g (a ) = 0 we
obtain g( a) = 0, oo that the degree of g(x)
> s. Hence f(x) is irre-
ducible, and the lemma is established.
To complete the proof of the theorem, let cu 1 , w 2 ,
... ,
w t be a gen-
erating system for the vector space E over F. Let f 1( x) be the separable
polynomial having w i as a root. Then E is the splitting field of
p(x) = f 1 (x)-f 2 (x)· .... f,(x).
If f(x) is a polynomial in a field F, and E the splitting field of
f (x ), then we shall call the group of automo:rphisms of E over F the
group of the equation f(x)
= 0.
We come now to a theorem known in
algebra as the Fundamental Theorem of Galois Theory which gives the
relation between the structure of a splitting field and its group
of automo:rphisms.
THEXJREM 16. (Fundamental Theorem). If p(x) is a separable
polynomial in a field F, and G the group of the equation p(x) = 0
where E is the
47
splitting field of p(x), then: (1) Each intermediate field, B,is the
fixed field for a subgroup G 8 of G, and distinct subgroups have distinct fixed fields. We say B and G 8 "belong" to each other. (2) The
intermediate field B is a normal extension of F if and only if the sub-
group G8 is a normal subgroup of G. In this case the group of automorphisms of B which leaves F fixed is isomorphic to the factor group
( G/G ). (3) For each intermediate field B, we have ( B/F) = index of
G 8 and ( E/B) = order of G,.
The first pm1 of the theorem cornes from the observation that E
is splitting field for p(x) when p(x) is taken to be in any intermediate
field. Hence, E is a normal extension of each intermediate field B, so
that B is the fixed field of the subgroup of G consisting of the automorphisms which leave B fixed. That distinct subgroups have distinct fixed
fields is stated in Corollary 2 to Theorem 14.
Let B be any intermediate field. Since B is the fixed field for
the subgroup G8 of G, by Theorem 14 we have ( E/B) = order of G,.
Let us call o(G) the order of a group G and i(G) its index. Then
o(G) "o(G 8 ) ·i(G
8
).
But (E/F) = o(G), and (E/F) = (E/B)·(B/F)
from which ( B/F) " i (G, ), which proves the third part of the theorem.
The number i( G 8
)
is equal to the number of lefr
cosets of G,.
The elements of G, being automorphisms of E, are isomorphisms of B;
that is, they map B isomorphically into some other subfield of E and
are the identity on F. The elements of G in any one coset of GB map B
in the same way. For let
aG 8 • Since a and a
1
2
a· a1 and a· a 2 be two elements of the coset
leave B fixed, for each a in B
48
we have aa 1 ( a)= a( a)= aa 2 ( a). Elements of different cosets give
different isomorphisms, for if a and r give the same isomorphism,
o(a)
=
r(a) for each a in B, then a· 1 r(a):::: a for each a in B. Hence,
1
0 - ,
:::
a 1 , where a is an element of G
rG 9
"
aa 1 G 9 ~ aG 9
1
8
.
But then r:::: aa 1 and
so that a and r belong to the same coset.
Each isomorphism of 8 which is the identity on F is given by an
automorphism belonging to G. For let
a be an isomorphism mapping 8
on B' and the identity on F. Then under a, p ( x) corresponds to p ( x ),
and E is the splitting field of p ( x) in 8 and of p( x) in 8 ' , By
Theorem 10, a can be extended to an automorphism a' of E, and since
a
I
leaves F fixed it belongs to G. Therefore, the number of distinct
isomorphisms of 8 is equal to the number of cosets of G 8 and is there-
fore equal to ( B/F).
The field a8 onto which a maps 8 has obviously aG 9 a-l as corresponding group, since the elements of a8 are left invariant by
precisely this group.
If B is a normal extension of F, the number of distinct automor-
phisms of 8 which leave F fixed is ( B/F) by Theorem 14. Conversely,
if the number of automorphisms is ( B/F) then B is a normal extension,
because if F 1 is the fixed field of all
these automorphisms,
then
F C F' C B, and by Theorem 14, ( B/F ')is equal to the number of
automorphisms in the group, hence ( 8/F ') = ( 8/F ). From ( B/F) =
(B/F')(F'/F) we have (F'/F) = 1 or F = F'. Thus, B is a normal
extension of F if and only if the number of automorphisms of B is ( B/F ).
B is a normal extension of F if and only
if each isomorphism of
8 into E is an automorphism of B. This follows from the fact that each
of the above conditions are equivalent to the assertion that there are
49
the same numberof isomorphisms and automorphisms. Since, for
each
a, B = aB is equivalent to aG 8 a -l C G 8 , we can finally say that B is
a normal extension of F and only if G 8 is a normal subgroup of G.
As we have shown, each isomorphism of B is described by the
effect of the elements of some left
coset of G,.
If B is a normal exten-
sion these isomorphisms are all automorphisms, but in this case the
cosets are elements of the factor group ( G/G
8
).
Thus, each automor-
phism of B corresponds uniquely to an element of ( G/G 8
versely. Since multiplication in ( G/G 8
)
)
and con-
is obtained by iterating the
mappings, the correspondence is an isomorphism between ( G/GB ) and
the group of automorphisms of B which leave F fixed. This completes
the proof of Theorem 16.
I. Finite Fields.
It is frequently necessary to know the nature of a finite subset
of a field which under multiplication in the field is a group. The
answer to this question is particularly simple.
THEOREM 17. If S is a finite subset (f 0 ) of a field F which
is a group under multiplication in F, then S is a cyclic group.
The proof is based on the following lemmas for abelian groups.
f.emma 1. If in an abelian group A and B are two elements of
orders a and b, and if c is the least common multiple of a and b, then
there is an element C of order c in the group.
50
Proof: (a) If a and b are relatively prime, C = AB has the required order ab. The order of C a = Ba is b and therefore c is divisible
by b. Similarly it is divisible by a. Since Cab = 1 it follows c = ab.
(b) If d is a divisor of a, we can find in the group an element of
order d. Indeed
Aa/d
is this element.
(c) Now let us consider the general case. Let pl' p 2 ,
... ,
Pr be
the prime numbers dividing either a or b and let
n
· · · Pr r
Call ti the larger of the two numbers ni and
c = P1
t
1
P2
t2
11\. Then
tr,
···
P, ·
n-
According to (b) we can find in the group an element of order pi 1 and
. one of order pi
one of order pi~ . Thus there 1s
'• . Part (a) shows that
the product of these elements will have the desired order c.
_Lemma 2:_. If there is an element C in an abelian group whose
_?rder c
i~
maximal (as is always the case if the group is finite) then c
is divisible by the order a of every element A in the group; hence
~c
1 is_ satisfied by each element in the group.
Proof: If a does not divide c, the greatest common multiple of a
and c would be larger than c and we
could find an element of that order,
thus contradicting the choice of c.
We now prove Theorem 17. Let n be the order of S and r the
largest order occuring in S. Then xr - 1 ::;:: 0 is satisfied for all ele-
51
ments of S. Since this polynomial of degree r in the field cannot have
more than r roots, it follows that r :2 n. On the other hand r
_s
n be-
cause the order of each element divides n. S is therefore a cyclic
group consisting of 1, f,
( 2,, .
where fn = 1.
, fn-1
Theorem 17 could also have been based on the decomposition
finite
theorem for abelian groups having a
number of generators. Since
this theorem will be needed later, we interpolate a proof of it here.
Let G be an abelian group, with group operation written as t-.
The element g , .
. ,
1
gk will be said to generate G if each element g of
, gk, g = n 1 g1 +
G can be written as sum of multiples of g,
If no set of fewer than k elements generate G, then g,
called a minimal generating system.
Any
.
+ nkgk.
, ~ will be
group having a finite genera-
ting system admits a minimal generating system. In particular, a finite
group aLways admits a minimal generating system.
From the identity n 1 ( g 1
it follows that if g 1 , g 2 ,
g,2, , .. gk
1
+ mg,) +
(n 2 .. n 1 m)g 2 = n 1g 1 + n 2 g 2
, gk generate G, also g 1 + mg,
generate G.
An equation m1 g 1 + m 2 g 2
+
+ mkgk = 0 will be called a re-
lation between the generators, and m 1 , .
. , mk will be called the co-
efficients in the relation.
We shall say that the abelian group G is the direct product of its
subgroups G 1 , G ,.
2
sumg = x
1
+ x2
t
. , Gk if each g
f
G is uniquely representable as a
+ x k' where xi
f
G , i = 1, , , . , k.
1
52
Decomposition Theorem. Each abelian group having a finite number of generators is the direct product of cyclic subgroups G 1 ,
where the order of G i divides the order of G i"'"l, i = 1,
the number of elements in a minimal generating system.
may each be infinite, in which case, to be
O(GJIO(G.
'
Gn
.
n-1 and n is
Gr, Gr+1 ,
Gn
precise,
)fori= 1,2, ... ,r-2).
+
We assume the theorem true for all groups having minimal generating systems of k-1 elements. If n = 1 the group is cyclic and the
theorem trivial. Now suppose G is an abelian group having a minimal
generating system of k elements. If no minimal generating system
fies a non-trivial relation, then let g , g ,
1
2
satis-
, gk be a minimal generating
system and G , G , . . . , Gk be the cyclic groups generated by them.
1
For each g
t
2
G, g = n g +
1 1
+ nkgk where the expression is
unique; otherwise we should obtain a relation. Thus the theorem would
be true. Assume now that some non-trivial relations hold for some minimal generating systems. Among all relations between minimal generating systems, let
(1)
be a relation in which the smallest positive coefficient occurs. After
an eventual reordering of the generators we can suppose this coefficient
to be m 1 , In any other relation between g ,
1
,
g,.
(2)
we must have m /n
1
1
•
Otherwise n 1 = qm 1
+ r, 0 < r < m1 and q times
relation (1) subtracted from relation (2) would yield a relation with a
coefficient r
< m1 .
Also in relation (1) we must have m/mi' i = 2,.
, k.
53
For suppose m
1
m
qm
2
1
+
1
<r <
r, 0
g1 + g2 , g2 ,
ID 1( g
does not divide one coefficient, say m2
••• ,
rn . In the generating system
1
gk we should have a relation
+ qg 2 ) + rg 2 + m3 g 3 +
r contradicts the
Then
choice
of m 1 •
+ mkqk = 0 where the coefficient
Hence
q2 m1 ,
m2 =
m3 =
q3m 1 ,
, gk is minimal generating, and m1 g-1 = 0. In any relation 0 = n 1if1 + n 2 g 2 +
g1 ,
since m 1 is a coefficient in a relation between
vious argument yields m1 n 1
,
and hence n 1
g·1 =
g 2,
g1
of
G1
and
G'. Each
(n
1
-
n;Jg1
-r-
(g'-
)g1
By
our inductive hypothesis, G
=0, so that
n1 [
1
n;g 1
=
cyclic groups generated by elements
. . . , tk
. , gk
.
and
G1 the
Then G is the direct product
satisfy ti
titl,
g1
+ g1
= n
•g1
1
+ g" implies
g") = 0, hence
n;
orders t 2 ,
gk our pre-
element g of G can be written
The representation is unique, since n 1
the relation (n 1
nkgk
0.
Let G 1 be the subgroup of G generated by g2 ,
cyclic group of order m generated by
1
+
l
andalso g'=
is the direct product of k-1
g2, g3 ,
,
i = 2,
ment applied to the generators g 1 , g 2 ,
.
g<~.
gk
whose respective
k-1. The preceding argugk
yields m 1 t 2 , from which
the theorem follows.
By a finite field is meant one having only a finite number
of elements.
Corollary. The non-zero elements of a finite field form a cyclic
If a is an element of a field F, let us denote the n-fold of a, i.e,,
54
the element ofF obtained by adding a to itself n times, by na. It is ob-
vious that n·(m·a)= (nm)·a and(n-a)(m·b) = nm·ab. If for one
element a
f. 0, there is an integer n such that n. a = 0 then n. b = 0
for each bin F, since n. b
=
(n.a)(a- 1 b)
--=
Q.a- 1 b
=
O.. Ifthere is a
positive integer p such that p · a = 0 for each a in F, and if p is the
smallest integer with this property, then F is said to have the charac-
1eristic_p. If no such positive integer exists then we say F has characteristic 0. The characteristic of a field is always a prime number. for if
P= r.sthenpa = rs.a=r.(s.a).However, s.a=b ,fOil a f,O
-f
and r b -,_; 0 since both r and S are less than
p,
to the definition of the characteristic. If na
0 for a ~ 0, then p divides
n, for n = qp
+
r where 0
<r < p
so that pa
and na = (qp
+
0 contrary
r)a = qpa + ra.
Hence na = 0 implies ra = 0 and from the definition of the characteristic
since r
< p,
we must have r = 0.
If F is a finite field having q elements and E an extension of F
, w , •. ,w n is
1
2
a basis of E over F, each element of E can be uniquely represented as
such that (E/F) = n, then E has qn elements. For if
a linear combination x 1 w 1 + x
F. Since each xi
ble
choices of x 1 ,
2
w2 +
f-
{iJ
xnwn where the x 1 belong to
can assume q values in F, there are qn distinct possi.
. , X
0
and hence qn distinct elements of E. E is
finite, hence, there is an element a of E
so that E = F(a). (The non-
zero elements of E form a cyclic group generated by a).
If we denote by P _ [ 0, 1, 2 , ... 1 p-1] the set of multiples of the
unit element in a field F of characteristic p, then P is a subfield of F
having p distinct elements. In fact, P is isomorphic to the field of
integers reduced mod p. If F is a finite field, then the degree of F over
55
Pis finite, say (F/P} = n, and F contains p 0 elements. In other
words, the order of any finite field is a power of its characteristic.
IfF and F' are two finite fields having the same order q, then
by the preceding, they have the same characteristic since q is a power
of the characteristic. The multiples of the unit in F and F 1 form two
fields P and P 1 which are isomorphic.
The non-zero elements ofF and F' form a group of order q-1
and, therefore, satisfy the equation
xq-l
1
~
0. The fields F and F
1
are splitting fields of the equation x q-1 = 1 considered as lying in P
and
p 1 respectively. By Theorem 10, the isomorphism between P and
P can be extended to an isomorphism between F and F ' . We have thus proved
THEOREM 18. Two finite fields having the same number of elements are isomorphic.
Differentiation. If f(x) = a o + a x + . . . + a nxn is a poly1
nomial in a field F, then we define f 1 ~a,
+ 2a 2 x + ... +
nanx n-l
The reader may readily verify that for each pair of polynomials f and
g we have
(f+g)'=f+g'
( f g)'
fg' + gf
(fn)'
THEOREM 19. The polynomial f has repeated roots if and only
if in the splitting field E the polynomials f and f' have a common
root. This condition is eauivalent to the assertion that f and ft have a
56
common factor of degree greater than 0 in F.
If a is a root of multiplicity k of f(x) then f ~ (x-a )"Q (x) where
~
Q(a)
f
1= (
x-<t
0. This gives
)kQ 1 ( , )
+ k ( x-a) k- 1 Q ( , ) ~ ( x-a) k-1
[ ( ,_, )
Q' ( , ) + kQ ( x)].
If k > 1, then a is a root of f' of multiplicity at least k-1. If
k
~
1, then f'(x)
~
Q(x) + (x-a)Q'(x) and f(a)
~
Q(a)
~
0. Thus,
f and f' have a root a in common if and only if a is a root off of
multiplicity greater than 1.
Iff and f' have a root a in common then the irreducible polynomial
in F having a as root divides both f and f' . Conversely,
any root of a
factor common to both f and f' is a root off and f ' .
Corollary. If F is a field of characteristic 0 then each irreducible
polynomial in F is separable.
Suppose to the contrary that the irreducible polynomial f(x) has
a root a of multiplicity greater than 1. Then, f
1
(x) is a polynomial
which is not identically zero (its leading coefficient is a multiple of
the leading coefficient of :flx) and is not zero since the characteristic
is 0) and of degree 1 less than the degree of f(x). But a is also a root
of f
(x) which contradicts the irreducibility of f(x).
J I_!oots_ of Unity.
If F is a field having any characteristic p, and E the splitting
field of the polynomial x "- 1 where p does not divide t:t, then we
shall refer to E as the field generated out ofF by the adjunction of a
primitive nth root of unity.
The polynomial x n
its derivative, nx n-l
,
1 does not have repeated roots in E, since
has only the root 0 and has, therefore, no roots
57
in common with x n- 1. Thus, E is a normal extension cf F.
If tl' t: 2
,,.,, €
0
are the roots of x n- 1 in E, they form a group under
multiplication and by Theorem 17 this group will be cyclic. If
1 ,f, f 2, ... , f n-1 are the elements of the group, we shall call t a primitive n
th
root of unity. The smallest power of
f
which is 1 is then
th,
THEOREM 20. If E is the field generated from F by a primitive
nth root of unity, then the group G of E over F is abelian for any n and
cyclic if n is a prime number.
We have E ~ F(E),
Thus, if
a and
since
the roots of
X0 -
is an integer 1 ::no
"r
f:.
rare distinct elements of G, a(£)
rootofx" -1 and, hence, a power off, Thus,
f
1 are powers of f,
< n.
r(f). But rJ(r) is a
a(t)=
Moreover, ra(i)::: r( ~
00
)::::
f
"awhere "a
(r(f)) "a:::
"a= ar(f), Thus, nar == q,n 7 mod n. Thus, the mapping of a on no
is a homomorphism of G into a multiplicative subgroup of the integers
mod n. Since r
;1:.
a implies r(f)
~
a(£), it follows that r f:- a implies
"a ~ n, mod n. Hence, the homomorphism is an isomorphism. If n is a
prime number, the multiplicative group of numbers forms a cyclic group.
K. Noether Equations.
If E is a field, and G = (
any set of elements
x0 ,
x, ,
a, r, .
in E will be said to provide a solution to
Noether's equations if xa. a ( x 1
element x
0
. ) a group of automorphisms of E,
) :::
x
= 0 then x = 0 for each r
1
for each a and r in G. If one
07
f
G. As
all values in G, and in the above equation x
07
T
:::
traces G, ar assumes
0 when x
in any solution of the Noether equations no element
0
xa
= 0. Thus,
= 0 unless the
solution is completely trivial. We shall assume in the sequel that the
58
trivial solution has been excluded.
THEOREM 21. The system x , x , ... is a solution to Noether's
equations if and only if there exists an element
Xa
a in E,
such that
ala( a ) for each a.
For any a, it is clear that xa =a/a( a) is a solution to the
equations, since
a!<>(a)·<>(a/r(a)) = a/a(a)·a(a)/OT(a) = a/m(a).
Conversely, let x
x , ... be a non-trivial solution. Since the
"' r
automorphisms a, r, ... are distinct they are linearly independent, and
the equation xa 'U(Z) + x, r ( z) + . . . = 0 does not hold identically.
Hence, there is an element a in E such that
xa""(a) + x,r(a) + ... = a~ 0. Applying
a(a) =
Multiplying by
a to
a
gives
L a(x,).ar(a).
uG
'lr gives
Xa·a(a) =
Replacing xa . a ( x, ) by
k xaa(x,).m(a).
uG
'trr
and noting that
ar assumes
all values in
G when r does, we have
Xa •<J(a) = L x,r(a) = a
nG
so that
xa = a!a(a).
A solution to the Noether equations defines a mapping C of G
into E, namely, C(a) = xa· If F is the fixed field of G, and the elements x lie in F, then C is a character of G. For
0
C(ar)= x01 = xa·a(x,) = xaxr= C(a)·C(r)sincea(x,) = x, if
x
7
f
F. Conversely, each character C of G in F provides a solution
59
to the Noether equations. Call C(o)
xo. Then, since
xr
f
F,
we have a( xr) = Xr . Thus,
Xa
·a ( x,) = xa · x, = C (a) · C ( r) = C( or)
xar.
We thereforehave,
by combining this with Theorem 21,
THEOREM 22. If G
is
over
the group of the normal field E
F,
then for each character C of G into F there exists an element a in E
such that C(a) = a/a(a) and, conversely, if a/a(
a)
is in F for each
a, then C (a)= a/a( a ) is a character of G. If r is the least common
multiple of the orders of elements of G, then a' E F.
We have already shown all but the last sentence of Theorem 22.
To prove this we need only show a(ar) =a' for
a'/a(a')
=
each a
f
G. But
(a/a(a))' = (C(a))' = C(a') = C(l) = 1.
L. Kummer's Fields.
IfF contains a primitive nth root of unity, any splitting field E
of a polynomial ( xn - a
i = 1,2,. .
1
)(
xn - a ) . . . ( xn - a ) where a. ( F for
2
'
'
, r will be called a Kummer extension of F, or more briefly,
a Kummer ..-field.,
If a field F contains a primitive nth root of unity, the number n
is not divisible by the characteristic of F. Suppose, to the contrary, F
has characteristic p and n :::: qp. Then yP _ 1 = (y - 1 )P since in the
expansion of (y - 1 )P each coefficient other than the first and last is
divisible by p and therefore is a multiple of the p-fnld of the unit of F
and thus is equal to 0. Therefore xn - 1 = (x q)p - 1 = ( xq - 1 )P
and
X
0
-
1 cannot have more than q distinct roots. But we assumed
that F has a primitive nth root of unity and 1, t,
t 2, . . . , (
n-1 would be
60
n distinct roots of xn - 1. It follows that n is not divisible by the
characteristic of F. For a Kummer field E, none of the factors
x" - a 1, a f 0 has repeated roots since the derivative, nxn·l , has
1
only the root 0 and has therefore no roots in common with xn - ai.
Therefore, the irreducible factors of xn - a, are separable, so that E
is a normal extension of F.
Let a 1 be a root of xn- ai in E. If
tinct
nth
roots of unity in F, then
roots of xn- ai' and
E
=
F(a 1 , a 2 ,
••• ,
f
1
, f
2
, . . . , t:
aifl' ait , . . . ,
2
0
are then dis-
a 1t:n will ben distinct
hence will be the roots of xn -a, so that
ar). Let a and r be two automorphisms in the group
G of E over F. For each ai' both a and r map ai on some other root of
x" - ai. Thus r(ai) =
f
17 a 1
and a(a) =
f iO
ai where fia and fir are
nth
roots of unity in the basic field F. It follows that
r(a(a,)) = r(<,aa,) =
'•ur(a)
= 'ia'lT a,= a(r(a,). Since oand r
are commutative over the generators of E, they commute over each element of E. Hence, G is commutative. If a
c.1 aa.,
a 2 (a.)
1
1
fia
01
=f.
2 a.,
10,
f
G, then a (a i)
=
01
etc. Thus, u (a.)
=a.1 for n.1 such that
1
= 1. Since the order of an n th root of unity is a divisor of n, we
have ni a divisor of n and the least common multiple m of n 1 , n 2 , •.. , n r
is adivisorofn. Since a"(u 1 ) = ai fori= 1,2, ... , r it follows thatm
is the order of a. Hence, the order of each element of G is a divisor of
n and, therefore, the least common multiple r of the orders of the elements of G is a divisor of n. If
fn/r
f
is a primitive n1h root of unity, then
is a primitive r th root of unity. These remarks can be summarized
in the following.
61
_THEXJREM 23. If Eisa Kummer field, i.e., a splitting field of
p(x) = (x"- a 1 )(x"- a 2 )
contains a primitive
nth
•.•
(x"- a,) whete a, lie in F, and F
root of unity, then: (a) E is a normal extension
ofF; (b) the group G of E over F is abelian, (c) the least common multi-
ple of the orders of the elements of G is a divisor of n.
Corollary._ If E is the splitting field of xP - a, and F contains a
primitive p th root of unity where p is a prime number, then either E ::: F
and xP -- a is split in F, or xP ... a is irreducible and the group of E
over F is cyclic of order p.
The order of
each element of G is, by Theorem 23, a divisor of p
and, hence, if the element is not the unit its order must be p. If
root of xP - a, then a, m, . . . , £p-I a are
all
a is
a
the roots of xP .... a so that
F(a ) = E and ( E/F) <]' . Hence, the order of G does not exceed p oo
that if G has one element different from the unit, it and its powers must
constitute all of G. Since G has p distinct elements and their behavior
is determined by their effect on a, then a must have p distinct images.
Hence, the irreducible equation in F for a must be of degree p and is
therefore
xP ...
a ::: 0.
The properties (a), (b) and (c) in Theorem 23 actually characterize
Kummer fields.
Let us suppose that E is a normal extension of a field F, whose
group G over F is abelian. Let us further assume that F contains a
primitive r th root of unity where r is the least common multiple of the
orders of elements of G.
The group of characters X of G into the group of r
th
roots of
62
unity is isomorphic to G. Moreover, to each a
a character C
f
f
G, if a
f:- 1,
there exists
X such that C (a) /:. 1. Write G as the direct product of
the cyclic groups G, G
2
, .
m,. Each
Gt of orders m 1 m 2
af G
Call C i the character sending a i
into ~',-, a primitive m . 'h root of unity and a. into 1 for j ~ i. Let C be
•
anyc h aracter. C ( ai )
Conversely, C
J
::::£i
1
11 •,ten
h
we h ave C = C 1 ''
C 1-lt
fLi •
t
I
defines
.
C I'
vl
Thus C
a
1
(
1/ 2
2
.
,
vt
C t I''
i
a character. Since the order of C. is
•
m 1, the character group X of G is isomorphic to G. If
a= a- 1
2
a f:-
1, then in
at least one vi, say v , is not divisible by m 1 .
.a 1
1
a)~ , v jo 1.
1
1
Let A denote the set of those non-zero elements
a of
E for which
a' c F and let F denote the non-zero elements of F. It is obvious that
1
A is a multiplicative group and that F1 is a subgroup of A. Let N denote the set of
rth
powers of elements in A and F~ the set of r th powers
of elements of F 1 . The following theorem provides in most applications
a convenient method for computing the group G.
THEOREM 24. The factor groups (A/F ) and (A'/F ')are isomorphic to each other and to the groups G and X.
We map A on A' by making a t A correspond to ar (A'. If ar
f F~,
where a E F then b E A is mapped on fl if and only if br = ar, that is,
1
if b is a solution to the equation xr ... ar = 0. But a, ca, c 2a, ... , (r-1 a
are distinct solutions to this equation and
since
f and a belong to
it follows that b must be one of these elements and must belong to
Thus, the inverse set in A of the subgroup F~ of N is F
1
factor groups (
A/F 1
)
and ( N IF~ ) are isomorphic.
,
F1,
F1 .
so that the
63
If a is an element of A, then (a/a(a))r = ar/u(ar) = 1. Hence,
a/a( a) is an
root of unity and lies in F
r'h
1 • ~Theorem
22, a/a (a)
defines a character C (a) of Gin F. We map a on the corresponding
character C. Each character C is by Theorem 22, image of some
Moreover,
a.a
1
is mapped on the character C
a.
* (a) =
a· a'/a(a·a') = a·a'!a(a)·a(a') = C(a)·C'(a) = C.C'(a), so
that the mapping is homomorphism.
The kernel of this homomorphism
is the set of those elements a for which a/u(a) = 1 for each
is F 1 • It follows, therefore, that ( A/F 1
also to G. In particular, (A/F 1
)
)
0 ,
hence
is isomorphic to X and hence
is a finite group.
We now prove the equivalence between Kummer fields and fields
satisfYing (a), (b) and (c) of Theorem 23.
THEDREM 25. If E is an extension field over F, then E is a
Kummet field if and only if E is normal, its group G is abelian and F
contains a primitive
rth
root
f
of unity where
r
is the least common
multiple of the orders of the elements of G.
The necessity is already contained in Theorem 23.. We prove the
sufficiency. Out of the group A, let
a1 F 1 , a 2 F 1,
... ,
atF1 be the co sets
ofF 1 . Since ai (A, we have af = ai (F. Thus, ai is a root of the
equation xr- a.=Oand sincem.,c 2a., . . . ,
'
'
'
fr-la.,
are also roots,
xr - ai must split in E. We prove that E is the splitting field of
(x'- a 1 )(x'- a 2
) •..
(x'- a,) which will complete the proof of the
theorem. 'lb this end it suffices to show that F ( a , a , . . . , at) =E.
1 2
64
, a,)
Suppose that F ( a , a , .
1
2
/c E.
Then F(a,
intermediate field between F and E, and since E is normal over
F (a,, .
, at) there exists an automorphism a c G,
leaves F (a 1 ,
•.. ,
f-
1, which
a,) fixed. There exists a character C of G for which
a
C(a) -/:. 1. Finally, there exists an element
C(o) ~ a/a(a) ~ 1. But a'
Moreover, A c F(a,
in F(a,
(1
f
in E such that
F 1 by Theorem 22, hence a, A.
,a, ) since all the cosets a iF1 are
,a,). Since F(a,
contained
, at) is by assumption left fixed by
a, a( a) = a which contradicts a/a(a) f- 1. It follows, therefore, that
F{a,, ... ,a 1 ) =E.
Corollary. If E is a normal extension of F, of prime order p, and
_ifF
contains a primitive pth root of unity, then E is splitting field of
an irreducible polynomial xP ... a in F.
a1 ,
E is generated by elements
,
an where af
not in F . Then xP - a is irreducible, for otherwise F(a
1
F. Let a 1 be
would be an
intermediate field between F and E of degree less than p, and by the
product theorem for the degrees, p would not be a prime number, contrary to assumption. E :::; F ( a 1
)
is the splitting field of
xP -
a.
M. Simple Extensions.
We consider the question of determining under what conditions
an extension field is generated by a
single element, called a primitive.
We prove the following
THEOREM 26. A
finite extension E of F is primitive over F if
65
~x_j!
there are only a finite number of intermediate fields.
(a) Let E
a in F.
~
F(a) and call f(x)
~
0 the irreducible equation for
Let B be an intermediate field and g(x) the irreducible equa-
tion for a in B. The coefficients of g(x) adjoined to F will generate a
field B
1
between F and B. g ( x ) is irreducible in B, hence also in B
Since E = B'(a) we see ( E/B )
~
~
( E/B ' ). This proves B'
1•
B. So B
is uniquely determined by the polynomial g(x). But g(x) is a divisor
of :flx), and there are only a finite number of possible divisors of :flx)
in E. Hence there are only a finite number of possible B's.
(b) Assume there are only a finite number of fields between E
and F. Should F consist only of a finite number of elements, then E is
generated by one element according to the Corollary on page 53. We
may therefore assume F to contain an infinity of elements. We prove:
To any two elements a, (3 there is a y in E such that F (a, (3 ) ~ F (y).
Let y :::a+ a(3 with a in F but for the moment undetermined. Con-
sider all the fields F(y) obtained in this way. Since we have an
infinity of a's at our disposal, we can find two, say a, and a 2 , such
that the corresponding y's, y1 ~
the same field F ( y
1
) ~
F ( y
2
).
a
+
aJ3
and y, ~ a +
a2 (3,
Since both y and y are in F ( y
1
2
yield
1
),
their difference (and therefore (3) is in this field. Consequently also
y1
-
a1
f3
~a.
So F(a,(3) C F(y 1 ). Since F(y 1 ) C F(a,(3) our con-
tention is proved. Select now
~
in E in such a way that ( F (
as large as possible. Every element
could find an element
proves E
~
F(1).
€
~
)IF ) is
of E mustbe in F( 1J) or else we
osuch that F(6) contains both ~and '·This
66
THEOREM 27. If E = F(a, a 2 ,
the field F, and a
~xists
1
a primitive
,
. . . , an)
is a finite extension of
a 2, ... , a 0 are separable elements in E, then there
ein E
such that E = F (e).
Proof: Let fi ( x) be the irreducible equation of
ai
in F and let B
be an extension of E that splits fi(x) 12 ( x) ... f,(x). Then B is normal
over F and contains, therefore, only a finite number of intermediate
fields (as many as there are subgroups of G).
8:)
the subfield E contains
only a finite number of intermediate fields. Theorem 26 now compietes
the proof.
N. Existence of a Normal Basis.
The following theorem is true for any field though we prove it
only in the case that F contains an infinity of elements.
fHE_OREM 28. If E is a normal extension ofF and a 1 , a , ... , a
are the elements of its group G, there is an element
& in E
such that
then elements a,( e), a,( 0)' ... , an( 0) are linearly independent with
respect to
F.
According to Theorem 27 there is an a such that E = F(u). Let
f(x) be the equation for a, put o,(a) = a i,
f(x)
f(x)
g(x) = (x-a)f'(a) and g,(x) = o,(g(x)) = -(x-q,)f'(q,)
g i( x ) is a polynomial in E having ak as root fork
(1)
g /X) gk (X) = 0 (mod f(x)) for i
I= i and hence
t
k.
In the equation
(2)
g*(x)
+
g 2 (x)
+ ... +
g,(x)- 1 = 0
the left side is of degree at most n - 1. If (2) is true for n different
values of x, the left side must be identically 0. Such n values are
67
apa 2 ,
•••
,a,,
since gi( ai) = 1 and gk(ai) c:::O fork-/::. i.
Multiplying (2) by g,( x) and using (I) shows:
(g,(x))'
(3)
g,(x) (mod f(x)).
=
We next compute the determinant
D(x) = 1a,ak(g(x)) I i,k = 1, 2, ... , n
(4)
and prove D(x) ~ 0. If we square it by multiplying column by column
and compute its value (mod f(x)) we get from (1), (2), (3) a determinant that has 1 in the diagonal and 0 elsewhere.
s0
(D(x)) 2 =I (mod f(x)).
D (x ) can have only a finite number of roots in F. Avoiding them
we can find a value a for x such that D(a) /:. 0. Now set (J = g(a).
Then the determinant
(5)
Consider any linear relation
t
ing the automorphism
xnan( ()) = 0 where the xi are in F. Apply-
a. to it would lead to n homogeneous equations for
'
the n unknowns xi. (5) shows that xi= 0 and our theorem is proved.
0. Theorem on Natural Irrationalities.
Let F be a field, p(x) a polynomial in F whose irreducible factors
are separable, and let E be a splitting field for p(x). Let B be an arbitrary extension of F, and let us denote by EB the splitting field of p(x)
when p(x) is taken to lie in B. If rx 1 ,
EB, then F(a,
.
, a 6 are the roots of p(x) in
,as) is a subfield of EB which is readily seen to
form a splitting field for p(x) in F. By Theorem 10, E and F(a 1 , .
a
5
)
68
are isomorphic. There is therefore no loss of generality if in the
we take E ::: F ( a 1 ,
••• ,
ofEB.Also,EB = B (a
sequel
a,) and assume therefore that E is a subfield
1
Let us denote by E
, ... ,
n
a,).
B the intersection of E and B. It is readily
seen that E n B is a field and is intermediate to F and E.
THEOREM 29. If G is the group of automorphisms of E over F,
_?.nd H
th~
group of EB over B, then H is isomorphic to the subgroup of
G having E n B as its fixed field.
Each automorphism of EB over B simply permutes al' ... , a,. in
some fashion and leaves B, and hence also F, fixed. Since the elements of EB are quotients of polynomial expressions in a , . . , as with
1
coefficients in B, the automorphism is completely determined by the
permutation it effects on a
1
, •
• ,
as, Thus,
each automorphism of EB
over B defines an automorphism of E = F ( a 1 ,
fixed. Distinct automorphisms, since a 1 , •
.. ,
a
5
)
which leaves F
, a 8 belong to E, have
different effects on E. Thus, the group H of EB ovet B can be considered as a subgroup of the group G of E over F. Each element of H
leaves E n B fixed since it leaves even all of B fixed. Howevet, any
element of E which is not in E
n
B is not in B, and hence would be
moved by at least one automorphism of H. It follows that E n B is the
fixed field of H.
~orollaty:.
If, undet the conditions of Theorem 29, the gtoup G is of
_prime on!er, then either H = G or H consists of the unit element alone.
69
III APPLICATIONS
by
A. N. Milgram
A. _Solvable_ Groups.
Before proceeding with the applications we must discuss certain
questions in the theory of groups. We shall assume several simple propositions:
f(x) =
xN
(a) If N is a normal subgroup of the group G, then the mapping
is a homomorphism of G on the factor group GIN. f is called
the natural homomorphism. (b) The image and the inverse image of a
normal subgroup under a homomorphism is a normal subgroup. (c) I f f
is a homomorphism of the group G on G' , then setting N' = f(N), and
defining the mapping g as g( xN ) = f(x) N
a homomorphism of the factor group
1 ,
we readily see that g is
G/N on the factor group G 1/N 1 •
Indeed, if N is the inverse image of N' then g is an isomorphism.
We now prove
THEOREM 1. (Zassenhaus). If U and V are subgroups of G, U and
v normal subgroups of U and V, respectively, then the following three
factor groups are isomorphic:
u (U nV) /u (U nv),
v(UnV)/v(unV), (UnV)/(unV)(vnU).
It is obvious that U
n
v is a normal subgroup of U
n
V. Let f
be the natural mapping of U on U/u. Call f(UnV) = H and f(Unv) = K.
Then f-'(H)
=
u(UnV) and C 1(K) = u(Unv) from which it follows
that u( UnV)/u( Unv) is isomorphic to H/K. If, however, we view f as
defined only over U n V, then f-\K) = [un(UnV)](Unv) =
(unV)(Urw) so that (UnV)/(unV)(Unv) is also isomorphic to H/K.
70
Thus the first and third of the above factor groups are isomorphic to
each other. Similarly, the second and third factor groups are isomorphic.
Corollary 1. If H is a subgroup and N a normal subgroup of the
group
G, then H/HnN is isomorphic to HN/N, a subgroup of G/N.
Proof:
Set G
= U, N = u, H
= V and the identity 1 ::::
v in
Theorem 1.
Corollary 2. Under the conditions of Corollary 1, if
GIN
is
_abelian, _so also is H/HnN.
Let us call a group G solvable if it contains a sequence of subgroups G = G0
_)
G1
.J, .. )
Gs = 1, each a normal subgroup of the
preceding, and with G i-1 ./G.1 abelian.
THEOREM 2. Any subgroup of a solvable group is solvable. For
let H be a subgroup of G, and call H 1 = HnG 1• Then that Hi_/Hi is
abelian follows from Corollary 2 above, where Gi-l' Gi and Hi-l play
the role of G, N and H.
THEOREM 3. The homomorph of a solvable group is solvable.
Let f(G) = G' , and defineG
1
i
=f( Gi)where Gi belongs to a
a sequence exhibiting the solvability of G. Then by (c) there exists a
homomorphism mappingGi_/Gi on G~_/Gf. But the homomorphic image
of an abelian group is abelian so that the groups G ~ exhibit the
solvability of G'
B. _Permutation Groups.
Any one to one mapping of a set of n objects on itself is called
a permu_tation: The iteration of two such mapping is called their product.
71
It
may be readily verified that the set of all such mappings forms a
group in which the unit is the identity map. The group is called the
symmetric group on n letters.
Let us for simplicity denote the set of n objects by the numbers
L 2,
... , n. The mapping S such that S(i)
noted by (123 ... n) and more generally (i j
ping T such that T ( i)
i + 1 mod n
be de-
. m) will denote the map-
j, ... , T ( m) ~ i. If ( i j ... m) has k numbers,
then it will be called a k cycle. It is clear that if T = (i j
T- 1
will
s) then
= (s . . ji).
We now establish the
Lemm.::: If a subgroup U of the symmetric group on n letters
(n > 4) contains every 3-cycle, and if u is a normal subgroup of U
such that U/u is abelian, then U contains every 3-cycle.
Proof: Let f be the natural homomorphism f(U) ,.... U/u and let
X
= (ijk), y
(krs) be twoelements of U, where i, j, k, r, s are 5
numbers. Then since U/u is abelian, setting f(x) = X • , f(y) = y
we have
1
f ( x-ly-1 xy) = X •-1 y H X 1 y 1 = 1, so that x- 1y- 1 xy ( u. But
x-ly-1xy= (kji)·(srk)·(ijk)·(krs) = (kjs) and for each k, j, s we
have (kjs) : u.
THEOREM 4. The symmetric group G on n letters is not solvable
for n > 4.
If there were a sequence exhibiting the solvability, since G con-
tains every 3-cycle, so would each succeeding group, and the sequence
could not end with the unit.
72
C. Solutjon of Equations by Radicals.
The extension field E over F is called an extension by radicals
if there exist intermediate fields 8 1 , B2
, ... ,
Br = E and Bi = Bi_ 1( a)
where each ai is a root of an equation of the form x ni
-
a. "" 0,
'
a 1 ' Bi-1 . A polynomial f ( x ) in a field F is said to be solvable by
radicals if its splitting field lies in an extension by radicals. We assume
unless otherwise specified that the base field has characteristic 0 and
that F contains as many roots of unity as are needed to make our subsequent statements valid.
Let us remark first that any extension ofF by radicals can always
be extended to an extension ofF by radicals which is normal over F.
Indeed B is a normal extension of B 0 since it contains not only a ,
1
butm
1
,
1
where (is any n 1 -root of unity, so that 8 1 is the splitting field
of x"t- a,. If f (x) ~ TT(x"2- a( a )), where a takes all values in
1
2
(J
the group of automorphisms of B 1 over B 0 , then f 1 is in B 0 , and
joining successively the roots of
xn 2 -
a( a
2
)
ad-
brings us to an exten-
sion of B which is normal over F. Continuing in this way we arrive at
2
an extension of E by radicals which will be normal over F. We now
prove
THEOREM 5. The polynomial f(x) is solvable by radicals if
and only if its group is solvable.
Suppose f(x) is solvable by radicals. Let E be a normal extension of F by radicals containing the sphtting field B of f(x), and call
G the group of E over F. Since for each i, Bi is a Kummer extension of
Bi-1, the group of Bi over Bi-t is abelian. In the sequence of groups
73
= Gil
='
=
J ... :J G 8 ,
1 each is a normal subgroup of the
1
preceding since GB
is the group of E over Bi·l and Bi is a normal
G
G8
i-1
extension of
Bi-l.
But GB _{G 8 . /s the group of (\ over B1_1 and hence
1
is abelian. Thus G is solvable. However, GB is a normal subgroup of
G, and G/GE is the group of B over F, and is therefore the group of the
polynomial f(x). But G/G 8 is a homomorph of the solvable group G and
hence is itself solvable.
On the other hand, suppose the group G of f(x) to be solvable
and let E be the splitting field. Let G = G 0 J G 1 :J . . . :J G, = 1 be a
sequence with abelian factor groups. Call B 1 the fixed field for G 1 •
Since Gi-l is the group of E over B 1_1 and Gi is a normal subgroup of
Gi·l' then Bi is nmmal over B1_1 and the group Gi_/Gi is abelian. Thus
Bi is a Kummer extension of Bi-t' hence is splitting field of a polynomial
of the fonn ( x"~•
1
)(
x"-a
2
) ••• (
splitting fields of the x n -
ak
x"-a,) so that by forming the successive
we see that Bi is an extension of B 1_1 by
radicals, from which it follows that E is an extension by radicals.
Remark. The assumption that F contains roots of unity is not
necessary in the above theorem. For if f(x) has a solvable group G,
then we may adjoin to F a primitive nth root of unity, where n is, say,
equal to the order of G. The group of f(x) when considered as lying in
F' is, by the theorem on Natural hrationalities, a subgroup G 1 of G,
and hence is solvable. Thus the splitting field over F
1
of f(x) can be
obtained by radicals. Conversely, if the splitting field E over F of f(x)
can be obtained by radicals, then by adjoining a suitable root of unity
E is extended to E 1 which is still normal over F
1
•
But E 1 could be
74
obtained by adjoining first the root of unity, and then the radicals, to
F; F would first be extended to F
E
1
•
1
Calling G the group of E
we see that G
1
1
and then F ' would be extended to
over F and G
is solvable and G/G
1
1
1
the group of E
is the group ofF
1
over F
1
,
over F and
hence abelian. Thus G is solvable. The factor group G/GE is the
group of f(x) and being a homomorph of a solvable group is
also solvable.
D. The General Equation of Degree n.
If F is a field, the collection of rational expressions in the
variables ul' u 2 ,
... , U0
with coefficients in F is a field F ( u
1
,
u2 ,
. . . , 11 0
By the general equation of degree n we mean the equation
f(x) = x"- u 1xn-l + u 2x"- 2 - + . . . + ( -1 )"u•.
(1)
Let E be the splitting field of f(x) over F(u 1 , u 2 ,
v 1 , v .2,
ul
••. , V
0
un). If
are the roots of f(x) in E, then
= VI + v2 + . . . + vn, u2 = vl v2 + vl v3. +
tVn-lVn'''
We shall prove that the group of E over F ( u 1 , u 2 ,
symmetric
•.. ,
... , U
0
)
is the
group.
Let F ( x,, x 2 ,
••• ,
xn) be the field generated from F by the
variables xl, x2, ...• xn. Let al = xt + x2 + ... + xn,
a2
=
x 1 x2 + x 1 x
3
-·
.. ,
+
X
0
_
1
x0
, ...
,an:: x 1 x 2 • . • xn be the ele-
mentary symmetric functions, i.e., (x-x ,)(x-x 2 ). .
xn- a xn-I +- ... (-l)"an = f*(x). If g(a 1 ,a 2 , .
1
polynomial in a 1 ,
...
,a,, then g(a 1 ,a 2 ,
...
(x-x,)
.,a,) is a
,a,)= 0 only if g is the
).
75
Ixi, Lx 1xk,
zero polynomial. For if g(
, ) = 0, then this relation
would
hold also if the xi were replaced by the vi. Thus,
g(Lvi, Lvi
0 or g(u 1 , u 2 ,
=
vk, ... )
... ,
un) = Ofromwhichitfollows
that g is identically zero.
Between the subfield F(a,
F(u
1
,
u
2
, •.. , U 0
we set up the following correspondence: Let
)
f(u 1 , , , , , U 0 )/g(u 1 ,,
f (a 1 ,
this correspond to
ping of F ( u
f(a 1 ,a 2 ,
•• ,
u
,
1
2
1 U0
, • .
)beanelementofF(u 1 ,
,
a0
) /
g (a 1'
Un) on all of F ( a 1 ,
,
,a 0 )/g(a 1 ,a 2 ,
•••
•.
,U
0
).Wemake
,a,). This is clearly a map,a,). Moreover, if
,an)
0. But this
, an), then fg, - gf,
= £1 (at,a:!• ... ,an)/gl(al,a2'.
implies by the above that
f(u 1 ,
...
,un)·g 1 (u 1 ,
..•
,un)-g(u 1 ,
so that f(ut' ... , un)/g(ut' u 2 ,
= f ( U , .•• ,
1
1
li
0
) /
g 1(
•.. ,
, ...
,un).f 1 (u 1 ,
..
,un)= 0
u.)
U , U , .•. , lln ).
1
2
themappingof F(u 1 ,u 2
...
It follows readily
from
this that
,un) on F(a 1 ,f1 2 , . . . ,an)is an isomor-
phism. But under this correspondence f(x) corresponds to f * ( x).
Since E and F(x 1 , x 2 ,
.•• ,
xn) are respectively splitting fields of f(x)
and f * ( x ), by Theorem 10 the isomorphism can be extended to an
iso-
morphism between E and F ( x 1 , x , . . , xn ). Therefore, the group of E
2
over F( u 1 , u , ... , un) is isomorphic to the group ofF ( x , x , ... , xn)
2
1 2
over F(a 1 ,a 2 ,
•.
,a,).
Each permutation of xl' x 2 ,, .. , xn leaves al'a 2 ,
and, therefore,
, an fixed
induces an automorphism ofF( XI' x 2 , .
leaves F(a 1,a , ... , a,.,) fixed. Conversely, each automorphism of
2
F ( x , x , . , , , xn ) which leaves F(a 1 , . . , a, ) fixed must permute the
1 2
roots x , x,., ... , x of f*(x) and is completely determined by the
1 "
n
76
permutation it effects on x 1 , x 2 , , , • 1 xn. Thus, the group of F ( x 1 , x 2 , ... , xn)
over F(a 1 ,a 2 ,
...
,an) is the symmetric group on n letters. Because of
the isomorphism between F ( x 1 ,
F ( u 1 , u2 ,
.•• ,
, , , 1
xn ) and E, the group for E over
un) is also the symmetric group. If we remark that the
symmetric group for n > 4 is not solvable, we obtain
from the theorem
on solvability of equations the famous theorem of Abel:
THEOREM 6. The group of the general equation of degree n is
the symmetric group on n letters. The general eguation of degree n is
not solvable by radicals if n > 4.
E. Solvable Equations of Prime Degree.
The group of an equation can always be considered as a permu-
tation group. If f (X) is a polynomial in a field F, let al' a,' . ' '' an be
the roots of f(x) in the splitting field E
~
F( a 1 , , . , , a 0
),
Then each
automorphism of E over F maps each root of f(x) into a root of f(x),
that is, permutes the roots. Since E is generated by the roots of f ( x ),
different automorphisms must effect distinct permutations. Thus, the
group of E over F is a permutation group acting on the roots
a 1 ,a 2 ,
...
,a,
of f(x).
For an irreducible equation this group is always transitive. For
let
a
and
a'
be any two roots of f(x), where f(x) is assumed irreduci-
ble. F(a) and F(a ')are isomorphic where the isomorphism is the
identity on F, and this isomorphism can be extended to an automorphism
of E (Theorem 10). Thus, there is an automorphism sending any given
root into any other root, which establishes the "transitivityn of the group.
77
A permutation
~tion_modulo
that o(i)
= bi
a of the numbers 1, 2, , .. , q is called a linear
q if there exists a number b
~
0 modulo q such
+ c(mod q), i = 1, 2, . . . ,q.
JHEOREM 7. Let f( x ) be an irreducible equation of prime de~ in
a field F. The group G of f( x) (which is a permutation group
of the roots, or the numbers 1, 2, ... , q) is solvable if and only if,
after a suitable change in the numbering of the roots, G is a group of
linear substitutions modulo q, and in the group G all the substitutions
withb = 1, a(i)
=
c + l(c = 1, 2 , ... , q) occur.
Let G be a transitive substitution group on the numbers
1, 2, ... , q and let G 1 be a normal subgroup of G. Let 1, 2, ... , k be the
images of 1 under the permutations of G 1 ; we say: 1, 2, ... , k is a
domain of transitivity of G,. If i <_q is a number not belonging to this
domain of transitivity, there is a
a f G which maps 1 on i. Then
a{ 1, 2, ... , k) is a domain of transitivity of aG 1 a -1 . Since G1 is a
normal subgroup of G, we have G 1
:::
aG 1a· 1 • Thus, u( 1, 2, ... , k) is
again a domain of transitivity of G 1 which contains the integer i and
has k elements. Since i was arbitrary, the domains of transitivity of
G 1 all contain k elements. Thus, the numbers 1, 2, ... , q are divided
into a collection of mutually exclusive sets, each containing k elements, so that k is a divisor of q. Thus, in case q is a prime, either
k = 1 (and then G 1 consists of the unit alone) or k :::: q and G 1 is
also transitive.
To prove the theorem, we consider the case in which G is
solvable. Let G = G 0 "J G 1 "J . . . J G s+l ~ 1 be a sequence exhibiting
the solvability. Since Gs is abelian, choosing a cyclic subgroup of it
78
would permit us to assume the term before the last to be cyclic, i.e.,
Gs is cyclic. If a is a generator of Gs, a must consist of a cycle con-
taining all q of the numbers 1, 2,,. , , q since in any other case G,
would not be transitive [if o = ( lij ... m)( n .. , p).,. then the powers
of a would map 1 only into 1, i, j , , .. m, contradicting the transitivity of
G, ]. By a change in the number of the permutation letters, we
may
assume
a ( i)
~
i + 1 (mod q)
a"(i) =o i + c (mod q)
Now let
r be any element of G
, TaT -1
is an element of G,, say
of G,. 1
then rar- 1( j) = ab( j)
=j + b (mod
5
_
1
.
Since Gs is a normal subgroup
TaT - 1 ~
a
h.
Let r( i) = j or ,- 1 ( j) =
q). Therefore,
ra ( i) oc r ( i)
~
b ( mod q) or r (i tl )
setting r(O)
~
c, we have r(l)
=r ( i)
=c +
+ b for each i. Thus,
b, r(2) =or( 1) + b = c + 2b
and in general r(i):: c-'- ib (mod q). Thus, each substitution in G s~l
is a linear substitution. Moreover, the only elements of
leave no element fixed belong toG , since for each a
i such that ai
+b
'
=;
Gs-l
~
i (mod q) [take i such that (a-1) i
which
1, there is an
=-
b].
We prove by an induction that the elements of G are all linear
substitutions, and that the only cycles of q letters belong toG,. Suppose the assertion true of Gs-n. Let r
f Gs-n-l
and let
a be a cycle
which belongs to Gs (hence also to Gs-n ). Since the transform of a
cycle is a cycle, r -lOT is a cycle in G!1-n and hence belongs to Gn ,
Thus ,- 1m ::.
ab
for some b. By the argument in the preceding para-
graph, r is a linear substitution hi + c and if
T
itself does not belong to
Gs, then r leaves one integer fixed and hence is not a cycle of q elements.
1,
79
We now prove the second half of the theorem. Suppose G is a
group of linear substitutions which
c(i)
:::=
contains a subgroup N of the form
i + c. Since the only linear substitutions which do not leave
an integer fixed belong to N, and since the transform of a cycle of q
elements is
again
In each coset N
where a ::_ i
a cycle of q elements, N is a normal subgroup of G.
r where r(i) ="- bi
+ c. But a- 1 r(
r(i) o-bi and r• (i)
i)
=- b'i then
=
(bi
+ c the substitution
a- 1 r occurs,
+ c) - c ::_ bi. Moreover,
if
rr 1 (i) ~ bb'i. Thus, thefactorgroup
( G/N) is isomorphic to a multiplicative subgroup of the numbers
1,2, ... , q-1 mod q and is therefore abelian. Since (G/N) and N are
both abelian, G is solvable.
Corollary 1. If G is a solvable transitive substitution group on q • •
letters (q prime), then the only substitution of G which leaves two or
more letters fixed is the identity.
This follows from the
q and bi
+ c
=i
fact
that each substitution is linear modulo
(mod q) has either no solution (b :;::; 1, c
exactly one solution(b
1=-
1) unless b
= 1,
c
=-
f.
0) or
0 in which case the sub-
stitution is the identity.
Corollary 2. A solvable, irreducible equation of prime degree in
a field which is a subset of the real numbers has either one real root
or all its roots are real.
The group of the equation is a solvable transitive substitution
group on q (prime) letters. In the splitting field (contained in the field
of complex numbers) the automorphism which maps a number
into its
complex conjugate would leave fixed all the real numbers. By Corollary
80
1, if two roots are left fixed, then all the roots are left fixed, so that
if the equation has two real roots all its roots are real.
F. Ruler and Compass Constructions.
Suppose there is given in the plane a finite number of elementary
geometric figures, that is, points, straight lines and circles. We seek
to construct others which satisfy ce11ain conditions in terms of the
given figures.
Permissible steps in the construction will entail the choice of
an arbitrary point interior to a given region, drawing a line through two
points and a circle with given center and radius, and finally intersecting pairs of lines, or circles, or a line and circle.
Since a straight line, or a line segment, or a circle is determined
by two points, we can consider ruler and compass constructions as con-
structions of points from given points, subject to certain conditions.
If we are given two points we may join them by a line, erect a
perpendicular to this line at, say 1 one of the points and, taking the distance between the two points to be the unit, we can with the compass
lay off any integer n on each of the lines. Moreover, by the usual
method, we can draw parallels and can construct
m/n.
Using the two
lines as axes of a cartesian coordinate system, we can with ruler and
compass construct all points with rational coordinates.
If a, b, c, ... are numbers involved as coordinates of points which
determine the figures given, then the sum, product, difference and
quotient of
any
two of these numbers can be constructed. Thus, each
81
element of the field R( a, b, c, ... ) which they generate out of the
rational numbers can be constructed.
It is required that an arbitrary point is any point of a given region.
If a construction by ruler and compass is possible, we can always
choose our arbitrary points as points having rational coordinates. If we
join two points with coefficients in R( a, b, c, ... ) by a line, its equation will have coefficients in R( a, b, c, . . . ) and the intersection of two
such lines will be a point with coordinates in R( a, b, c, ... ). The equation of a circle will have coefficients in the field if the circle passes
through three points whose coordinates are in the field or if its center
and one point have coordinates in the field. However, the coordinates
of the intersection of two such circles, or a straight line and circle, will
involve square roots.
It follows that if a point can be constructed with a ruler and com-
pass, its coordinates must be obtainable from R( a, b, c, ... ) by a formula
only involving square roots, that is, its coordinates will lie in a field
R, J R,_ 1
field over
)
...
Ri-l
J R 1 = R(a, b, c, ... ) where each field R 1 is splitting
of a quadratic equation x 2
-
6, p. 21) since either R, = R 1_1 or ( R/R 1_1
a :::: 0. It follows (Theorem
)
= 2, that ( R,IR 1 ) is a
power of two. If x is the coordinate of a constructed point, then
(R 1 ( x)/R 1
) •(
R,/R 1 (x)) = (R,/R,) = zv oo that R 1 ( x)/R 1 must also
be a power of two.
Conversely, if the coordinates of a point can be obtained from
R( a, b, c, ... ) by a formula involving square roots only, then the point
can be constructed by ruler and compass. For, the field operations of
82
addition, subtraction, multiplication and division may be performed by
ruler and compass constructions and, also, square roots using 1: r =
r : r to obtain r :::
1
compass
v1 r 1 may be performed by means of ruler and
instructions.
As an illustration of these considerations 1 let us show that it is
impossible to trisect an angle of 604 Suppose we have drawn the unit
circle with center at the vertex of the angle, and set up our coordinate
system with X-axis as a side of the angle and origin at the vertex.
Trisection of the angle would be equivalent to the construction
of the point (cos
20°,
sin 20°) on the unit circle. From the equation
cos 30 = 4 cos3 0 - 3 cos 0, the abscissa would satisfy
4x 3
-
3x =
1/2.
The reader may readily verify that this equation has
no rational roots, and is therefore irreducible in the field of rational
numbers. But since we may assume only a straight line and unit
length given, and since the 60° angle can be
R (a, b, c,. .
constructed,
we may take
..) to be the field R of rational numbers. A root a of the
irreducible equation 8x3 - 6x - 1 ~ o is such that ( R (a )/R) ~ 3,
and not a power of two.